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In the proof of the Ratio Test. We assume the terms of the sum are all positive and we have $$\lim_{n \to \infty} \frac{a_{n+1}}{a_n} = L < r < 1$$

Then we say there is an $N$ such that $$\frac{a_{n+1}}{a_n} \leq r$$

when $n \geq N$

Why are we allow to have $\leq$? I am under the impression that this is what's going on

Since $\frac{a_{n+1}}{a_n} $ converges, then for any $\epsilon >0$, there is an $N$ such that $\forall n \geq N$, we have $$\left | \frac{a_{n+1}}{a_n} - L\right | < \epsilon \iff -\epsilon + L < \frac{a_{n+1}}{a_n} < L + \epsilon$$

Here we take $r = L + \epsilon$ and we are only looking at the right inequality.

Also, can someone write me a formula for a sequence that has a limit $L$, but goes over $1$ initially (or sometimes) and then goes near $L$ after a very long time (for large $N$)?

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Consider the sequence $a_n = k{1 + \sin n \over n}$, with $k > 1$ it will have value $>1$ initially, but approach $1$. Rational or integer sequences can be constructed which have similar features. – abiessu Aug 25 '13 at 19:52
Where are you reading this proof from? I am quite sure that $|{\frac{a_{n+1}}{a_n}}| < r$ (strictly less than $r$). – kvmu Aug 25 '13 at 20:55
Spivak's Calculus (3ed) – Hawk Aug 25 '13 at 20:57

For convergence, limit L from the ratio test should be less than 1. If we assume that this limit is r, then for convergence, r<1. Now for any finite n, L is the limit of convergence. And so with the assumption that r is the limit of our ratio, we establish L

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For some reason, I don't feel like you are answering my question at all... – Hawk Aug 25 '13 at 20:10
My answer was a bit edited, though correctly. I have tried to explain the proof I have from my references. If you wish, you could take a look at Stewart's Calculus, 7th edition, Chapter 11 section 8 THere you find exactly the proof you are looking for. I am sure other Calc books may also have it, don't know on top of my head. My answer is certainly not a formal proof, I readily admit; just tried to explain the jest of the proof. – imranfat Aug 25 '13 at 20:17
I have read the proof already, my question is regarding the proof. I am not asking "how to do the proof" – Hawk Aug 25 '13 at 20:40
In the last sentence you showed that the ratio is greater than 1 for n less than 6. That's not what the ratio test is about. The ratio test is what happens to the division of two successive values of a sequence when n goes to infinity. I suspect that's where the misconnect between you and the proof is. Otherwise honestly I wouldn't know. – imranfat Aug 25 '13 at 22:33

You are exactly right about the existence of $r$; you can take it to be $L+\epsilon$ as you say.

If you take the series $\sum_{n=1}^\infty (\frac{3}{4})^n(n+2)(n+1)$, for example, then

$\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=\lim_{n\to\infty}\frac{3}{4}\cdot\frac{n+3}{n+1}=\frac{3}{4}<1$; but $\frac{a_{n+1}}{a_n}\ge1$ for $n\le5$.

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So when (if any time) can we have $$\frac{a_{n+1}}{a_n} \leq r$$ for $r \in (3/4, 1)$? – Hawk Aug 25 '13 at 21:23
Since $\frac{a_{n+1}}{a_n}\le\frac{27}{28}$ for $n\ge6$, you could take $r=\frac{27}{28}$ and $N=6$, for example. (You could also take $r=\frac{7}{8}$ and $N=11$, say.) The main thing to remember is that for any choice of $r\in(3/4,1)$, you will be able to find a suitable $N$. – user84413 Aug 25 '13 at 22:18
But the property $$\frac{a_{n+1}}{a_n} \leq r$$ can't be true for every (positive) sequence? – Hawk Aug 25 '13 at 23:48
No, you're right; but if $\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=L<1$, then you can apply your argument above to get an $r$ and an $N$ with $\frac{a_{n+1}}{a_n}\le r$ for $n\ge N$ where $r<1$. – user84413 Aug 26 '13 at 0:06
So you are saying it is indeed always possible for any sequence with $$\lim_{n \to \infty} \frac{a_{n+1}}{a_n} = L < r <1$$. Then we may find an $N$ such that $\forall n \geq N$, we get $$\frac{a_{n+1}}{a_n} \leq r$$ I am not convinced that it is true for every $n \geq N$. Can't a sequence bounce back and forth a bit before convergence while being above and below $r$ sometimes? – Hawk Aug 26 '13 at 1:05

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