# Proving part of the ratio test

1. Jun 2, 2014

### mindarson

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

1. The problem statement, all variables and given/known data

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

2. Relevant equations

3. 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 ∑an converges absolutely, I need to show that ∑|an| 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

∑an = |a1| + |a2| + |a3| + ... + |an|

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

Supposing lim n→∞ |an+1/an| = L < 1, this means that there is some integer N such that for n > N, all the ratios |an+1/an| 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

∑|an| = |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 ∑|an| converges, ∑an 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!

2. Jun 2, 2014

### mindarson

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.

3. Jun 2, 2014

### 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.

4. Jun 2, 2014

### johnqwertyful

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

5. Jun 2, 2014

### BiGyElLoWhAt

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

$∑(5-\frac{n}{1000})$

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.

6. Jun 2, 2014

### BiGyElLoWhAt

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

7. Jun 2, 2014

### jbunniii

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 $n > N$,
$$|a_n| = \left|\frac{a_{n}}{a_{n-1}}\right|\cdots\left|\frac{a_{N+1}}{a_N}\right||a_N|$$

8. Jun 2, 2014

### pasmith

No, $5 - \frac{n}{1000} \to -\infty$. Also, the condition is that $$\lim_{n \to \infty} \frac{|a_{n+1}|}{|a_n|} < 1$$, and for $a_n = 5 - n/1000$ we get $$\frac{a_{n+1}}{a_n} = \frac{n + 1 - 5000}{n - 5000} \to 1$$ 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 $|a_n| \to a > 0$ then also $|a_{n+1}| \to a > 0$ 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 $0/0$.