Proving part of the ratio test

[mindarson]

This is not a homework problem. I'm doing it for fun. But it is the kind that might appear on homework.

Homework Statement

I'm trying to prove that if lim n→∞ |a_n+1/a_n| = L < 1, then \Sigma a_n converges absolutely and therefore converges.

Homework Equations

The Attempt at a Solution

Here's my thinking. I feel like I'm on the right track, but I may need some help formalizing my expression of what's happening "in the limit" as n→∞.

To show that ∑a_n converges absolutely, I need to show that ∑|a_n| converges. My strategy is to show that, because of the condition above, as n→∞ this "tends toward" a geometric series with common ratio < 1 and therefore converges.

I have

∑a_n = |a₁| + |a₂| + |a₃| + ... + |a_n|

= |a₁| + |a₁||a₂/a₁| + |a₁||a₂/a₁||a₃/a₂| + ... + |a₁||a₂/a₁||a₃/a₂||a₄/a₃|...|a_n/a_n-1|

Supposing lim n→∞ |a_n+1/a_n| = L < 1, this means that there is some integer N such that for n > N, all the ratios |a_n+1/a_n| are equal in the sense that the difference between any two of them can be made arbitrarily small by choosing N appropriately.

Therefore I conclude that for n > N, the series above can be written

∑|a_n| = |a₁|(1 + L + L² + L³ + L⁴ + ... + Lⁿ)

i.e. it is a geometric series (or eventually becomes one beyond N) with common ratio < 1 and therefore converges.

And finally, since ∑|a_n| converges, ∑a_n converges absolutely and therefore it also converges.

Is my reasoning solid here? I am particularly concerned about the conclusion that all the ratios in the series (the ones that multiply |a₁| in each term) eventually equal L. I think my intuition here is correct, but I am not well enough attuned to the subtleties of analysis to be confident that my chain of reasoning is unimpeachable.

NOTE: I know there must be better, shorter, more elegant, less cumbersome ways of proving this. I'm not interested in those until I have developed my own proof to the utmost. That way I will learn the most from the process.

Thanks!

 

[mindarson]

Supposing lim n→∞ |a_n+1/a_n| = L < 1, this means that there is some integer N such that for n > N, all the ratios |a_n+1/a_n| are equal in the sense that the difference between any two of them can be made arbitrarily small by choosing N appropriately.

This is what troubles me most. A bit hand-wavey, I fear. I am wondering if I shouldn't try to show this explicitly using epsilonic reasoning, etc.

 

[gopher_p]

Rather than showing your series "tends toward" a geometric series, maybe try to show that it is eventually smaller than a convergent geometric series.

 

[johnqwertyful]

In undergraduate analysis, most proofs require an epsilon. This is one of them.

 

[BiGyElLoWhAt]

hmmm... is that a proper representation of the ratio test?

What about something like this:

$∑(5-\frac{n}{1000})$

Or is this a situation where the ratio test doesn't apply? I'm just asking, because this fits within your conditions, $\frac{a_{n+1}}{a_{n}}<1$ but as n approaches infinity, $a_{n}→5$

I think you need:
if
$\frac{a_{n+1}}{a_{n}}<1$ and $\displaystyle\lim_{n→\infty}a_{n}=0$ then S is convergant.

I know this doesn't really help you answer your question, but I feel it's a necessary condition.

 

[BiGyElLoWhAt]

oh wait, I lied. $a_{n}→-\infty$
either way, it doesn't converge...

 

[jbunniii]

I think you need:
if
$\frac{a_{n+1}}{a_{n}}<1$ and $\displaystyle\lim_{n→\infty}a_{n}=0$ then S is convergant.

No, that's not enough to ensure convergence. If $a_n = 1/n$ then $a_n \rightarrow 0$
$$\frac{1/(n+1)}{1/n} = \frac{n}{n+1} < 1$$
but $\sum a_n$ diverges.

I suggest the following approach. We are given that
$$\lim \left|\frac{a_{n+1}}{a_n}\right| = L < 1$$
Therefore if we pick some number $M$ satisfying $L < M < 1$, then there is some $N$ such that
$$\left|\frac{a_{n+1}}{a_n}\right| < M$$
for all $n \geq N$. Now use the fact that if $n > N$,
$$|a_n| = \left|\frac{a_{n}}{a_{n-1}}\right|\cdots\left|\frac{a_{N+1}}{a_N}\right||a_N|$$

 

[pasmith]

hmmm... is that a proper representation of the ratio test?

What about something like this:

$∑(5-\frac{n}{1000})$

Or is this a situation where the ratio test doesn't apply? I'm just asking, because this fits within your conditions, $\frac{a_{n+1}}{a_{n}}<1$ but as n approaches infinity, $a_{n}→5$

No, 5 - \frac{n}{1000} \to -\infty. Also, the condition is that <br /> \lim_{n \to \infty} \frac{|a_{n+1}|}{|a_n|} &lt; 1, and for a_n = 5 - n/1000 we get <br /> \frac{a_{n+1}}{a_n} = \frac{n + 1 - 5000}{n - 5000} \to 1 so this is not a counter-example.

In fact, if |a_{n+1}|/|a_n| tends to anything other than 1 then it must be the case that |a_n| \to 0: If |a_n| \to a &gt; 0 then also |a_{n+1}| \to a &gt; 0 and thus |a_{n+1}|/|a_{n}| \to a/a = 1 by the theorem that if the limits of two sequences exist, then the limit of the product exists and is the product of the limits. But if |a_n| \to 0 then the sequence 1/|a_{n}| diverges to +\infty and that theorem doesn't apply: the limit of |a_{n+1}|/|a_{n}| is in this case of the indeterminate form 0/0.