Consider a random variable with mean 0 and variance 1 (this is like the "natural units" of second-moment probability). Now this variance puts a value on how dispersed the PDF of the random variable can be: one couldn't, for example, have two Dirac-delta poles really far from the origin, because you can calculate the variance of that, and it's too high.
Which raises the question: for some $k$, what's the maximum fraction of the distribution that can be outside $(-k,k)$?
Ok, one thing is clear: at this maximum, there should be nothing outside $[-k,k]$, because then you could bring inwards without changing the fraction but reducing the variance.
Also, for any value in $(-k,k)$, if you moved it over to the mean (0), the variance would go down. So the distribution must be comprised of just three poles at 0, $-k$ and $+k$, and is necessarily symmetric so that the mean is at 0. Letting $p/2$ be the height of each pole at $k$, the variance in terms of $p$ is $pk^2$. So $pk^2=1$, and $p=1/k^2$. I.e.
$$P\left(\left|X\right|>k\right) \le 1/k^2$$
Or for general mean and variance:
$$P\left(\left|\frac{X-\mu}{\sigma}\right|>k\right) \le 1/k^2$$
This is Chebyshev's inequality, and gives you a limit on how much of the distribution can be some given distance $k$ from the mean. Note how it only becomes interesting for large $k$ ($k>1$).
Well, clearly this approach seems to open up a whole world of similar inequalities. Another is the Markov inequality, which states that (for a nonnegative random variable) no more than $1/k$ of a population can have value more than $k$ times the mean, i.e.
$$P\left(X\ge k\right)\le\frac{\mu}{k}$$
(Justify this with similar reasoning as Chebyshev's.)
In fact, Chebyshev's inequality can be derived as a special case of Markov's (do it).
The point of these inequalities is that means and variances are generally easy to track, even when probability distributions are unknown. Providing bounds on probability fractions based on these is very useful for proving convergence in probability -- for example, the weak law of large numbers becomes elementary with Chebyshev's inequality.
Insights on interesting things.
Showing posts with label convergence. Show all posts
Showing posts with label convergence. Show all posts
Probabilistic convergence
The law of large numbers is something that we know, that in our heads is almost the definition of probability (it's not): "as sampling increases, the average of a variable approaches its expected value". I.e. for $X_i$ IID:
$$\lim_{n\to\infty}\frac1n \sum{X_i}=\mu$$
Let's think about what this statement really says: when you take more and more readings of $X$, the average will go closer and closer to $\mu$. But the values of these readings are inherently probabilistic: this is not an actual sequence of real numbers you can take the limit of. Rather, you are talking saying that of all the possible realizations (which are real number sequences), almost all of them (probabilistically) converge to the thing. I.e.
$$\mathrm{Pr}\left[\lim_{n\to\infty}X_n=X\right]=1$$
This is known as almost sure convergence.
In general, the thing on the right could've been a random variable, rather than a real number. And here's where some probability theory (read the article) comes in, because the random variables $X_n$ and $X$ need to be defined on the same sample space for this to make sense (i.e. it's not just about the distribution).
But with this, the definition as above still works: as an example, consider the sample space $[0,1]$ and consider a sequence of random variables $X_n$ that is respectively 1 on some corresponding sequence of sub-intervals approaching $[0,1/2]$. Then this approaches the random variable that is 1 on $[0,1/2]$ almost surely.
And yes, this is entirely due to the correlations between these things.
In any case, almost sure convergence isn't really the best way to express random variables converging to each other, as you can see. E.g. the central limit theorem -- like $\frac{1}{\sqrt{n}}\sum\frac{X_n-\mu}{\sigma}\sim N(0,1)$, cannot be phrased in terms of almost sure convergence, because $N(0,1)$ is a distribution, not a random variable.
Indeed, you may have figured that the problem of a random sequence converging to a random variable is somewhat similar to the notion of "functions converging to a function" -- indeed, one may think of the distributions of the random variables in the sequence and discuss their convergence. I.e.
$$F_n(x)\to F(x)$$
This is called convergence in distribution.
While convergence in distribution does not imply almost sure convergence in general as we've seen, we would expect that it does imply it in the case where the limiting random variable is constant (because then issue of correlations disappears).
But you may realize that this is not really so: a sequence may look increasingly like something without actually limiting to it. For example, think about a sequence like 1, 1, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1... with an infinite number of 1s, but decreasing in frequency. This doesn't limit to zero. If this were a deterministic sequence, this would never be expected to limit to 0 as the positions of the 1s would be hardcoded into the generation of the sequence. However, the sequence can also be realized as a realization of a sequence of random variables $X_n$ that have probability $1/n$ of being 1. Then the $X_n$ converge in distribution to 0, but their realizations almost never (thus in particular don't almost surely) converge to 0.
So it seems that asking for realizations to almost surely converge to the right thing is a bit too strong for a lot of purposes. A weaker notion of convergence than almost sure convergence can be constructed by considering probabilities of each $X_n$ separately rather than as a sequence: $X_n$ converges to $X$ if each $X_n$ is in the limit almost surely arbitrarily close to $X$. Or more precisely:
$$\lim_{n\to\infty}\mathrm{Pr}\left(\left|X_n-X\right|<\varepsilon\right)=1$$
This is known as convergence in probability. Indeed:
Exercise:
Prove Slutsky's lemma: given $X_n, Y_n$ converge to $X,y$ in probability and $y$ is a constant random variable:
$$\lim_{n\to\infty}\frac1n \sum{X_i}=\mu$$
Let's think about what this statement really says: when you take more and more readings of $X$, the average will go closer and closer to $\mu$. But the values of these readings are inherently probabilistic: this is not an actual sequence of real numbers you can take the limit of. Rather, you are talking saying that of all the possible realizations (which are real number sequences), almost all of them (probabilistically) converge to the thing. I.e.
$$\mathrm{Pr}\left[\lim_{n\to\infty}X_n=X\right]=1$$
This is known as almost sure convergence.
In general, the thing on the right could've been a random variable, rather than a real number. And here's where some probability theory (read the article) comes in, because the random variables $X_n$ and $X$ need to be defined on the same sample space for this to make sense (i.e. it's not just about the distribution).
But with this, the definition as above still works: as an example, consider the sample space $[0,1]$ and consider a sequence of random variables $X_n$ that is respectively 1 on some corresponding sequence of sub-intervals approaching $[0,1/2]$. Then this approaches the random variable that is 1 on $[0,1/2]$ almost surely.
And yes, this is entirely due to the correlations between these things.
In any case, almost sure convergence isn't really the best way to express random variables converging to each other, as you can see. E.g. the central limit theorem -- like $\frac{1}{\sqrt{n}}\sum\frac{X_n-\mu}{\sigma}\sim N(0,1)$, cannot be phrased in terms of almost sure convergence, because $N(0,1)$ is a distribution, not a random variable.
Indeed, you may have figured that the problem of a random sequence converging to a random variable is somewhat similar to the notion of "functions converging to a function" -- indeed, one may think of the distributions of the random variables in the sequence and discuss their convergence. I.e.
$$F_n(x)\to F(x)$$
This is called convergence in distribution.
While convergence in distribution does not imply almost sure convergence in general as we've seen, we would expect that it does imply it in the case where the limiting random variable is constant (because then issue of correlations disappears).
But you may realize that this is not really so: a sequence may look increasingly like something without actually limiting to it. For example, think about a sequence like 1, 1, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1... with an infinite number of 1s, but decreasing in frequency. This doesn't limit to zero. If this were a deterministic sequence, this would never be expected to limit to 0 as the positions of the 1s would be hardcoded into the generation of the sequence. However, the sequence can also be realized as a realization of a sequence of random variables $X_n$ that have probability $1/n$ of being 1. Then the $X_n$ converge in distribution to 0, but their realizations almost never (thus in particular don't almost surely) converge to 0.
So it seems that asking for realizations to almost surely converge to the right thing is a bit too strong for a lot of purposes. A weaker notion of convergence than almost sure convergence can be constructed by considering probabilities of each $X_n$ separately rather than as a sequence: $X_n$ converges to $X$ if each $X_n$ is in the limit almost surely arbitrarily close to $X$. Or more precisely:
$$\lim_{n\to\infty}\mathrm{Pr}\left(\left|X_n-X\right|<\varepsilon\right)=1$$
This is known as convergence in probability. Indeed:
- Almost sure convergence implies convergence in probability (obviously).
- Convergence in probability implies convergence in distribution (because they are both topological notions of convergence and the map from a random variable to its distribution is continuous).
- When the limit random variable is constant, convergence in distribution implies convergence in probability.
Exercise:
Prove Slutsky's lemma: given $X_n, Y_n$ converge to $X,y$ in probability and $y$ is a constant random variable:
- $X_n+Y_n$ converges to $X+y$ in probability.
- $X_nY_n$ converges to $Xy$ in probability.
- $X_n/Y_n$ converges to $X/y$ in probability.
Why is it necessary that $y$ be a constant?
What's with e^(-1/x)? On smooth non-analytic functions: part I
When you first learned about the Taylor series, your intuition probably went something like this: you have $f(x)$, the derivative at this point tells you how $f$ changes from $x$ to $x+dx$ (which tells you $f(x+dx)$), the second derivative tells you how $f'$ changes from $x$ to $x+dx$, which recursively tells you $f(x+2\ dx)$, the third derivative tells you $f(x+3\ dx)$, and so on -- so if you have an infinite number of derivatives, you know how each derivative changes, so you should be able to predict the full global behaviour of the function, assuming it is infinitely differentiable (smooth) throughout.
Everything is nice and dandy in this picture. But then you come across two disastrous, life-changing facts that make you cry for those good old days:
Everything is nice and dandy in this picture. But then you come across two disastrous, life-changing facts that make you cry for those good old days:
- Taylor series have radii of convergence -- If I can predict the behaviour of a function up until a certain point, why can't I predict it a bit afterwards? It makes sense if the function becomes rough at that point, like if it jumps to infinity, but even functions like $1/(1+x^2)$ have this problem. Sure, we've heard the explanation involving complex numbers, but why should we care about the complex singularities (here's a question: do we care about quaternion singularities?)? Specifically, a Taylor series may have a zero radius of convergence. Points around which a Taylor series has a zero radius of convergence are called Pringsheim points.
- Weird crap -- Like $e^{-1/x}$. Here, the Taylor series does converge, but it converges to the wrong thing -- in this case, to zero. Points at which the Taylor series doesn't equal a function on any neighbourhood, despite converging, are called Cauchy points.
In this article, we'll address the weird crap -- $e^{-1/x}$ (or "$e^{-1/x}$ for $x>0$, 0 for $x= 0$" if you want to be annoyingly formal about it) will be the example we'll use throughout, so if you haven't already seen this, go plot it on Desmos and get a feel for how it looks near the origin.
Terminology: We'll refer to smooth non-analytic functions as defective functions.
Terminology: We'll refer to smooth non-analytic functions as defective functions.
The thing to realise about $e^{-1/x}$ is that the Taylor series -- $0 + 0x + 0x^2 + ...$ -- isn't wrong. The truncated Taylor series of degree $n$ is the best polynomial approximation for the function near zero, and none of the logic here fails for $e^{-1/x}$. There is honestly no other polynomial that better approximates the shape of the function as $x\to 0$.
If you think about it this way, it isn't too surprising that such a function exists -- what we have is a function that goes to zero as $x\to 0$ faster than any polynomial does. I.e. a function $g(x)$ such that
$$\forall n, \lim\limits_{x\to0}\frac{g(x)}{x^n}=0$$
This is not fundamentally any weirder than a function that escapes to infinity faster than all polynomials. In fact, such functions are quite directly connected. Given a function $f(x)$ satisfying:
$$\forall n, \lim\limits_{x\to\infty} \frac{x^n}{f(x)} = 0$$
We can make the substitution $x\leftrightarrow 1/x$ to get
$$\forall n, \lim\limits_{x\to0} \frac{1}{x^n f(1/x)} = 0$$
So $\frac1{f(1/x)}$ is a valid $g(x)$. Indeed, we can generate plenty of the standard smooth non-analytic functions this way: $f(x)=e^x$ gives $g(x)=e^{-1/x}$, $f(x)=x^x$ gives $g(x)=x^{1/x}$, $f(x)=x!$ gives $g(x)=\frac1{(1/x)!}$ etc.
To better study what exactly is going on here, consider Taylor expanding $e^{-1/x}$ around some point other than 0, or equivalently, expanding $e^{-1/(x+\varepsilon)}$ around 0. One can see that:
$$\begin{array}{*{20}{c}}{f(0) = {e^{ - 1/\varepsilon }}}\\{f'(0) = \frac{1}{{{\varepsilon ^2}}}{e^{ - 1/\varepsilon }}}\\{f''(0) = \frac{{ - 2\varepsilon + 1}}{{{\varepsilon ^4}}}{e^{ - 1/\varepsilon }}}\\{f'''(0) = \frac{{6{\varepsilon ^2} - 6\varepsilon + 1}}{{{\varepsilon ^6}}}{e^{ - 1/\varepsilon }}}\\ \vdots \end{array}$$
Or ignoring higher-order terms for our purposes,
$$f^{(N)}(0)\approx(1/\varepsilon)^{2N}e^{-1/\varepsilon}$$
Each derivative $\frac{e^{-1/\varepsilon}}{\varepsilon^{2N}}\to0$ as $\varepsilon\to0$, but they each approach zero slower than the previous derivative, and somehow that is enough to give the sequence of derivatives the "kick" that they need in the domino effect that follows -- from somewhere at $N=\infty$ (putting it non-rigorously) -- to make the function grow as $x$ leaves zero, even though all the derivatives were zero at $x=0$.
But we can still make it work -- by letting $N$, the upper limit of the summation approach $\infty$ first, before $\varepsilon\to 0$. In other words, instead of directly computing the derivatives $f^{(n)}(0)$, we consider the terms
$$\begin{array}{*{20}{c}}{f_\varepsilon^{(0)} = f(0)}\\{{{f}_\varepsilon^{(1)} }(0) = \frac{{f(\varepsilon ) - f(0)}}{\varepsilon }}\\{{{f}_\varepsilon^{(2)} }(0) = \frac{{f(2\varepsilon ) - 2f(\varepsilon ) + f(0)}}{{{\varepsilon ^2}}}}\\{{{f}_\varepsilon^{(3)} }(0) = \frac{{f(3\varepsilon ) - 3f(2\varepsilon ) + 3f(\varepsilon ) - f(0)}}{{{\varepsilon ^3}}}}\\ \vdots \end{array}$$
And write the generalised Hille-Taylor series as:
$$f(x) = \mathop {\lim }\limits_{\varepsilon \to 0} \sum\limits_{n = 0}^\infty {\frac{{{x^n}}}{{n!}}f_\varepsilon ^{(n)}(0)} $$
Then $N\to\infty$ before $\varepsilon\to0$ so you "reach" $N\to\infty$ first (or rather, you get large $n$th derivatives for increasing $n$) before $\varepsilon$ gets to 0.
Another way of thinking about it is that the "local determines global" stuff makes sense to predict the value of the function at $N\varepsilon$, countable $N$, but it's a stretch to talk about uncountably many $\varepsilon$s away, which is what a finite neighbourhood is. But with these difference operators in the Hille-Taylor series, one ensures that each neighbourhood is a finite multiple of $h$ away at any point, so the differences determine $f$.
Very simple (but fun to plot on Desmos) exercise: use $e^{-1/x}$ or another defective function to construct a "bump function", i.e. a smooth function that is 0 outside $(0, 1)$, but takes non-zero values everywhere in that range.
Similarly, construct a "transition function", i.e. a smooth function that is 0 for $x\le0$, 1 for $x\ge1$. (hint: think of a transition as going from a state with "none of the fraction" to "all of the fraction")
If you're done, play around with this (but no peeking): desmos.com/calculator/ccf2goi9bj
Similarly, construct a "transition function", i.e. a smooth function that is 0 for $x\le0$, 1 for $x\ge1$. (hint: think of a transition as going from a state with "none of the fraction" to "all of the fraction")
If you're done, play around with this (but no peeking): desmos.com/calculator/ccf2goi9bj
Intuition to convergence
We've all seen these kinds of sums. You start with something obviously divergent, like:
$$S = 1 + 2 + 4 + 8 + ...$$
And then apply standard manipulations on it to obtain a bizarrely finite result:
$$\begin{gathered}
\Rightarrow S = 1 + 2(1 + 2 + 4 + ...) \hfill \\
\Rightarrow S = 1 + 2S \hfill \\
\Rightarrow S = - 1 \hfill \\
\end{gathered} $$
How exactly is this result to be interpreted? Surely the definition of an infinite sum is as a limit of a finite sum as the upper limit increases without bound -- by this definition it would seem that $S$ evidently doesn't approach $-1$, it diverges to infinity. Is there, then, something wrong with the form of our argument? And if so, why does it seem to work for so many other sums, like convergent geometric progressions?
We'll get to all that in a moment, but first, let's talk about how to fold a tie into thirds. We know how to fold a tie -- or a strip of paper or a rope or whatever -- into halves, into quarters, into any power of two. But how would one fold it into thirds? Sure, we can approximate it by trial and error, but is there a more efficient algorithm to approximate it?
Here's one way: start with some approximation to 1/3 of the tie -- any approximation, however good or bad. Now consider the rest of the tie (~2/3) and fold it in half. Take one of these halves -- this is demonstrably a better approximation to 1/3 than your original. In fact, the error in this approximation is exactly half the error in the original approximation. You can keep repeating this process, and approach an arbitrarily close value to 1/3.
Why does it work? Well, it's obvious why it works. More interestingly, how could one have come up with this technique from scratch?
The key insight here is that if you had started from exactly 1/3 and performed this algorithm, defined as $x_{n+1}=\frac12(1-x_n)$, the sequence would be constant -- it would be 1/3s all the way down.
However, this is not a sufficient argument. For instance, here's another sequence of which 1/3 is a fixed point: the algorithm $x_{n+1}=1-2x_n$. However here, if you were to start with any other number but 1/3, the sequence would not approach 1/3, but rather diverge away. While 1/3 is still a fixed point, this is an unstable fixed point, while in the previous case it was a stable fixed point.
But what exactly is wrong with extending the same argument to $x_{n+1}=1-2x_n$? Well, perhaps we should state the argument precisely in the case of $x_{n+1}=\frac12(1-x_n)$. The reason we know this converges to 1/3 regardless of the initial value is that 1/3 is the only value which stays the same in the algorithm (i.e. is a steady-state solution). Convergence of the sequence requires that the sequence the fluctuations get smaller, i.e. the sequence approaches a value that doesn't fluctuate around, it approaches a steady state.
But this reveals our central assumption -- we assumed that the sequence is convergent at all! If it is convergent, then 1/3 is the only value it could converge to, because convergence means approaching a steady state, and 1/3 is the only steady state.
The same principle applies to our original problem -- an infinite series is also a sequence, a sequence of partial sums. Our mistake is really in this step:
$$\begin{gathered}
... \hfill \\
1 + 2(1 + 2 + 4 + 8 + ...) = 1 + 2S \hfill \\
\end{gathered} $$
By declaring that this is the same $S$, we have assumed that this sum really has a value. To be even clearer, consider this (taking $n\to\infty$):
$$\begin{gathered}
S = 1 + 2 + 4 + ... + {2^n} \hfill \\
S = 1 + 2(1 + 2 + 4 + ... + {2^{n - 1}}) = 1 + 2S? \hfill \\
\end{gathered} $$
In other words, we assumed that $S$ reaches a steady state, that removing the last term $2^n$ wouldn't change the value of the summation. This would've been true if we were dealing with $(1/2)^n$ instead, because then the partial sum does reach a steady state, since its "derivative", $(1/2)^n$, approaches approaches 0.
With that said, the sum $1+2+4+8+...=-1$ (and other such surprising results) can in fact be correct. What we've proven here is that if the sum converges, it converges to -1. Otherwise, it's $2^\infty -1$. If you can construct an axiomatic system in which the sum does converge, where 0 behaves like $2^\infty$ in some specific sense, then the identity would be true. Such a system does in fact exist, it's called the 2-adic system.
You know, there is a sense in which you can understand the 2-adic system. When you take partial sums of $1+2+4+8+...$, you always get sums that are "1 less than a power of 2". $1+2+4+8+16=2^5-1$, for example -- what's the significance of $2^5$? Well, it's a number which 2 divides into 5 times. What's a number that 2 divides into an infinite number of times? Well, it's zero, and $0-1=-1$. Ths might sound like a ridiculous argument, and indeed it is false in our conventional algebra system, but the foundation of the 2-adic system.
The understanding of convergence we gained here -- from the tie example -- was pretty fantastic. It applies to all sorts of infinite sequences -- ordinary recurrences, (such in the form of) infinite series, continued fractions, etc. The idea of stable and unstable fixed points is a general one, and a very important one. Recommended watching:
$$S = 1 + 2 + 4 + 8 + ...$$
And then apply standard manipulations on it to obtain a bizarrely finite result:
$$\begin{gathered}
\Rightarrow S = 1 + 2(1 + 2 + 4 + ...) \hfill \\
\Rightarrow S = 1 + 2S \hfill \\
\Rightarrow S = - 1 \hfill \\
\end{gathered} $$
How exactly is this result to be interpreted? Surely the definition of an infinite sum is as a limit of a finite sum as the upper limit increases without bound -- by this definition it would seem that $S$ evidently doesn't approach $-1$, it diverges to infinity. Is there, then, something wrong with the form of our argument? And if so, why does it seem to work for so many other sums, like convergent geometric progressions?
We'll get to all that in a moment, but first, let's talk about how to fold a tie into thirds. We know how to fold a tie -- or a strip of paper or a rope or whatever -- into halves, into quarters, into any power of two. But how would one fold it into thirds? Sure, we can approximate it by trial and error, but is there a more efficient algorithm to approximate it?
Here's one way: start with some approximation to 1/3 of the tie -- any approximation, however good or bad. Now consider the rest of the tie (~2/3) and fold it in half. Take one of these halves -- this is demonstrably a better approximation to 1/3 than your original. In fact, the error in this approximation is exactly half the error in the original approximation. You can keep repeating this process, and approach an arbitrarily close value to 1/3.
Why does it work? Well, it's obvious why it works. More interestingly, how could one have come up with this technique from scratch?
The key insight here is that if you had started from exactly 1/3 and performed this algorithm, defined as $x_{n+1}=\frac12(1-x_n)$, the sequence would be constant -- it would be 1/3s all the way down.
However, this is not a sufficient argument. For instance, here's another sequence of which 1/3 is a fixed point: the algorithm $x_{n+1}=1-2x_n$. However here, if you were to start with any other number but 1/3, the sequence would not approach 1/3, but rather diverge away. While 1/3 is still a fixed point, this is an unstable fixed point, while in the previous case it was a stable fixed point.
But what exactly is wrong with extending the same argument to $x_{n+1}=1-2x_n$? Well, perhaps we should state the argument precisely in the case of $x_{n+1}=\frac12(1-x_n)$. The reason we know this converges to 1/3 regardless of the initial value is that 1/3 is the only value which stays the same in the algorithm (i.e. is a steady-state solution). Convergence of the sequence requires that the sequence the fluctuations get smaller, i.e. the sequence approaches a value that doesn't fluctuate around, it approaches a steady state.
But this reveals our central assumption -- we assumed that the sequence is convergent at all! If it is convergent, then 1/3 is the only value it could converge to, because convergence means approaching a steady state, and 1/3 is the only steady state.
The same principle applies to our original problem -- an infinite series is also a sequence, a sequence of partial sums. Our mistake is really in this step:
$$\begin{gathered}
... \hfill \\
1 + 2(1 + 2 + 4 + 8 + ...) = 1 + 2S \hfill \\
\end{gathered} $$
By declaring that this is the same $S$, we have assumed that this sum really has a value. To be even clearer, consider this (taking $n\to\infty$):
$$\begin{gathered}
S = 1 + 2 + 4 + ... + {2^n} \hfill \\
S = 1 + 2(1 + 2 + 4 + ... + {2^{n - 1}}) = 1 + 2S? \hfill \\
\end{gathered} $$
In other words, we assumed that $S$ reaches a steady state, that removing the last term $2^n$ wouldn't change the value of the summation. This would've been true if we were dealing with $(1/2)^n$ instead, because then the partial sum does reach a steady state, since its "derivative", $(1/2)^n$, approaches approaches 0.
With that said, the sum $1+2+4+8+...=-1$ (and other such surprising results) can in fact be correct. What we've proven here is that if the sum converges, it converges to -1. Otherwise, it's $2^\infty -1$. If you can construct an axiomatic system in which the sum does converge, where 0 behaves like $2^\infty$ in some specific sense, then the identity would be true. Such a system does in fact exist, it's called the 2-adic system.
You know, there is a sense in which you can understand the 2-adic system. When you take partial sums of $1+2+4+8+...$, you always get sums that are "1 less than a power of 2". $1+2+4+8+16=2^5-1$, for example -- what's the significance of $2^5$? Well, it's a number which 2 divides into 5 times. What's a number that 2 divides into an infinite number of times? Well, it's zero, and $0-1=-1$. Ths might sound like a ridiculous argument, and indeed it is false in our conventional algebra system, but the foundation of the 2-adic system.
Explain similarly why $1+3+9+27+...=-1/2$ in the 2-adic system.
The understanding of convergence we gained here -- from the tie example -- was pretty fantastic. It applies to all sorts of infinite sequences -- ordinary recurrences, (such in the form of) infinite series, continued fractions, etc. The idea of stable and unstable fixed points is a general one, and a very important one. Recommended watching:
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