Checking convergence 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?
 A: 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 $\lim a_n \leq \lim b_n$. 
A: 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!
A: 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.
A: The 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 the 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 than $1$ for convergence and not just the ratio.
