# Checking convergance of series $\sum_{n=1}^\infty \frac {3^{2n+1}}{9^n\sqrt n}$ using ratio test

So, I have the following series:

$\sum_{n=1}^\infty \frac {3^{2n+1}}{9^n\sqrt n}$

I want to check if it is convergent. I tried using the Ratio Test Method and what I got was the following limit:

$\lim_{n\to \infty}\frac{\sqrt n}{\sqrt {n+1}}$

Obviously, said limit is equal to 1, so the ratio test should be inconclusive. But I was wondering, why is it like that? As $\sqrt n < \sqrt{n+1}$ for all $n \in \mathbb N$, so technically, the absolute value of the ratio is always less than $1$, so the series should be convergent. What am I missing here?

• @VukBibic I think you should change your title. People are getting confused about what your question really is, and they are answering the wrong one. :P – Franklin Pezzuti Dyer Jun 29 '17 at 18:57

First of all, you should be checking the limit as $n\to\infty$, not as $n\to 0$.

Second of all, the whole point of a limit is that it finds the value approached, not each of the values reached along the way. Just because each ratio is less than one doesn't mean that they don't approach one. For example, take the following series: $$\sum_{n=1}^\infty \frac{n}{n+1}$$ It obviously diverges. But if we use the ratio test, we get that the ratio between two consecutive terms is $$\frac{n(n+2)}{(n+1)^2}$$ Which is always less than 1, so by your reasoning, this series is convergent, right?

No. It obviously is not convergent. Again, the entire concept of a limit is the value that is approached. Even though $e^{-x}$ is never $0$ for real $x$, $$\lim_{x \to \infty} e^{-x}=0$$ You might want review the concept of the limit. If anything else is confusing you, or if I have misunderstood your question, please let me know and I will be happy to revise my answer!

• Your answer is based on the idea that something that does not work in another example so it will not work here. It still doesn't explain the gap in his reasoning or addresses his concerns. – Anurag A Jun 29 '17 at 18:59
• @AnuragA I'm not basing my argument on that example, I'm just providing another example in which such a fallacy obviously leads to the wrong conclusion. My main point is that a limit is defined by the quantity approached by an expression. – Franklin Pezzuti Dyer Jun 29 '17 at 19:01
• @anuraga That is not correct. To show that the test is inconclusive, we need only show one example of a divergent series for which the ratio test yields a limit of $1$ and one convergent series for which the ratio test yields a limit of $1$. – Mark Viola Jun 29 '17 at 20:43
• @MarkViola I don't think OP is confused about the ratio test being inconclusive (if you read his sentence: "Obviously...."). His confusion is that even though all the terms of the sequence $\sqrt{\frac{n}{n+1}}$ are less than $1$ so how can one claim the limit is $1$? My remark is that just because there are other sequences which have the same behavior is not a mathematical justification to claim that the sequence here should also have the same behavior. – Anurag A Jun 29 '17 at 20:47
• Actually, I have just now read your answer a few more times and realised that it's not enough for the absolute value of the ratio to be less than one for a series to converge. Thank you a lot! :) – Vuk Bibic Jun 30 '17 at 0:20

Note that if $a_n <L$ for all $n \in \mathbb N$, then $\lim a_n \leq l$.

Similarly, if $a_n < b_n$ for all $n \in \mathbb N$. Then $\lim a_n \leq \lim b_n$.

You may want some justification for the general fact that $a_n < a$ for all $n$ implies $\color{blue}{\lim_{n\to\infty}a_n \le a}$, since that seems to be the source of at least part of your confusion; it is indeed possible for equality to hold, as in your very example of $\lim_{n\to\infty}\sqrt{\frac{n}{n+1}}$.

Of course, the only alternative to $\lim_{n\to\infty}a_n \le a$ is for $\lim_{n\to\infty}a_n > a$, so suppose this were so; call $\lim_{n\to\infty}a_n = L$.

Since $L > a$, rearrange to see that $L - a > 0$. Now, by the definition of convergence, there must be some positive integer $N$ such that for every positive integer $n\ge N$, $|a_n - L| < L-a$. Unpack this into the chain of inequalities $$-(L-a) < a_n - L < L - a \iff a - L < a_n - L < L-a\implies \color{red}{a < a_n\ \text{for all n\ge N,}}$$ which does contradict what we supposed that $a_n < a$ for all $n$.

Only one of $\lim_{n\to\infty}a_n \le a$, $\lim_{n\to\infty}a_n > a$ can be true, and the latter leads us to a contradiction, so the former must be true.

Absolute value of the ratio is always less than $1$, is not a sufficient condition for convergence of the series.

For a simple example, consider the simple example of a divergent series $\sum \frac1n$. As $n<n+1$, the absolute value of ratio is strictly less than $1$ even though the limit is equal to $1$.

You can go through the proof of ratio test to see it is important that the limit is strictly less that $1$ for convergence and not just the ratio.