Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $N$)?

share|improve this question
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

2 Answers 2

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

share|improve this answer
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 $n\le5$.

share|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.