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I'm trying to find a couple of examples — the simpler the better — of sequences $(a_n:n=1, 2, \dots)$ where you:

(i) assume the $\lim_{n \rightarrow \infty}a_n$ exists and then determine (this is the "guess" part) what that limit $L$ must be; and only afterward,

(ii) prove that, in fact, $\lim_{n \rightarrow \infty}a_n = L$ (this is the "check" part).

The kinds of example I'm looking for are those where one does not simply use standard "rules for limits" to do both steps at once, i.e. not something like $a_n = (n+1)/n$, where we would simply use $\lim_{n \rightarrow \infty} (n+1)/n = \lim_{n \rightarrow \infty} (1 + 1/n) = 1 + \lim_{n \rightarrow \infty} (1/n) = 1 + 0 = 1$.

Possibly such examples would be defined recursively. For example,$$a_1 = 1, \, \,a_n=\frac{1}{2}\left(a_{n-1}+\frac{2}{a_{n-1}}\right).$$ In this example, if you assume the sequence converges, with limit $L$, then necessarily $L = (1/2)\left(L+2/L\right)$ whence $L = \sqrt{2}$. However, unless I'm mistaken, to prove that $\lim_{n \rightarrow \infty} a_n=\sqrt{2}$ is not so easy.

Another such example where the "guess" part is easy is: $$a_1=1, \,\, a_n = \sqrt{2 + x_{n-1}},$$ but again I think the "check" part would have to proceed without exploiting the knowledge of what the guessed limit (2) must be.

For the "check" step (ii), I intend that one actually uses the knowledge from (i) of what the limit must be (having assumed that the limit exists), in order to go ahead and show that this is, in fact, the limit. Thus for (ii) I do not want to apply principles such as that an increasing sequence bounded above must converge, principles whose application would presumably be independent of what the limit must be.

Any such examples?

murray
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2 Answers2

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For your example with $L=\sqrt 2$, one way to prove the limit is to look at $a_n-\sqrt 2$ and to prove that $|a_{n+1}-\sqrt 2 \lt k |a_n-\sqrt 2|$ for some $k \lt 1$. This proves the limit is correct, but is very hard to do if you don't know the limit in advance, so I think it is just what you are looking for. This approach would apply to your second example as well.

Ross Millikan
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  • I don't see why $|a_{n+1}-\sqrt 2 \lt k |a_n-\sqrt 2|$ for some $k \lt 1$. I experimented (with Mathematica) and found that this seems to be true even for $k = (10^{100}-1)/10^{100}$ only for $n \leq 5$. (P.S. I'm aware that what's involved here is actually Newton's Method.) – murray Aug 10 '17 at 16:09
  • Yes, so when you are close the error is about squared each step. Just saying the error decreases is not enough for convergence to $L$. If we start out within $1/2$, say, we can prove the error decreases by at least a factor $3/4$ at every step, which is enough – Ross Millikan Aug 10 '17 at 17:08
  • are you sure? The initial value $a_1=1$ is certainly within $1/2$ of $\sqrt{2}$, but when I use 100-digit precision, I find that the error-decrease factor at $n = 6$ to $n = 7$ is not at most $3/4$. And even if I try 1000-digit precision, again that decrease factor exceeds $3/4$ from $n = 12$ to $n = 13$. – murray Aug 10 '17 at 18:11
  • Yes. Write $a_n=\sqrt 2+x$ where $x$ is small enough. Then $a_{n+1}=\frac 12(\sqrt 2 +x +\frac 2{\sqrt 2+x})\approx \sqrt 2+\frac {x^2}{2\sqrt 2}$ where I kept the highest term in the Taylor series that doesn't cancel, so as long as $x \lt \frac 12$ the error is multiplied by a factor less than $\frac 1{4\sqrt 2}$ – Ross Millikan Aug 10 '17 at 20:20
  • Taylor series of $\frac{1}{2}\left(\sqrt{2} + x + \frac{2}{\sqrt{2}+x}\right)$ as a function of $x$? And about what center of expansion -- about 0? (In any case, there's a "multitude of sins" disguised in that $\approx$, where one really needs an inequality!) – murray Aug 11 '17 at 16:28
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I suspect it will be hard to give examples of a sequence where we need to guess first and then only be able to prove using the guess (and not be able to apply other techniques).

For instance, a very similar sequence to what you have appeared in my answer here: Limit of recursive sequences: $x_{n+1}=\frac 1 3 (x_n+2y_n), y_{n+1}=\frac 1 3 \left( \frac 1 {x_n}+ \frac 1 {2y_n} \right).$

The sequence was

$$c_{n+1} = \frac{1}{3}\left(c_n + \frac{1}{c_n}\right)$$

The proof I had was to use that $c_n \to \frac{1}{\sqrt{2}}$ and in an answer similar to Ross' bound the error term geometrically.

Then in the comments it was pointed out that we don't really need that, since this is just $x_{n+1} = f(x_n)$ with $|f'(x)| \lt 1$.

So while I used the fact initially that $c_n$ should converge to $\frac{1}{\sqrt{2}}$, I didn't really need to use it. I suspect it is the same with your example sequence.

That said, this guess and prove technique was espoused by Donald J Newman as a good method to figure out the asymptotics of sequences. One such method for guessing:

Say the sequence was $$y_{n+1} = y_n + \frac{1}{y_n}$$

Look at this as

$$y_n(y_{n+1} - y_n) = 1$$

and write this as a differential equation

$$ y y' = 1$$

subsituting $y = y_n$ and $y' = y_{n+1} - y_n$

This gives us

$(y^2)' = 2$ and thus

$$ y^2 \sim 2n$$

Thus the guess we have is $$y_n \sim \sqrt{2n}$$

And we can indeed prove that. See my answer here: Given $a_{1}=1, \ a_{n+1}=a_{n}+\frac{1}{a_{n}}$, find $\lim \limits_{n\to\infty}\frac{a_{n}}{n}$

I suspect we can prove $y_n \sim \sqrt{2n}$ without guessing it first. Perhaps the approach in this answer can be used to do that: https://math.stackexchange.com/a/2383179/1102

Aryabhata
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  • But I am looking for (simple) examples where, if you can "guess" readily what the limit $L$ is (assuming it exists), then you can use your knowledge of what the value of $L$ is in order to prove convergence. – murray Aug 10 '17 at 20:28
  • @murray: The first sequence $c_n$ is exactly that. The linked answer proves convergence using the guess that the limit must be $L = \frac{1}{\sqrt{2}}$. It proves that $|c_{n+1} - L| \lt k |c_n - L|$ for some $0 \lt k \lt 1$. For the second example consider the sequence $\frac{y_n}{\sqrt{n}}$ instead. – Aryabhata Aug 10 '17 at 21:07