Convergent or divergent $\sum_{n=1}^{\infty}\frac{1}{\sqrt{n^2+1}}\left(\frac{n}{n+1}\right)^n$? Any suggestions? I have tried using D'Alembert's test, but on the end I get 1. I can't think of any other series with which to compare it. In my textbook the give the following solution which I don't quite understand:
$\sum_{n=1}^{\infty}\frac{1}{\sqrt{n^2+1}}(\frac{n}{n+1})^n=\sum_{n=1}^{\infty}\frac{1}{\sqrt{n^2+1}}\frac{1}{(1+\frac{1}{n})^n}\sim \sum_{n=1}^{\infty}\frac{1}{n}\frac{1}{e} \sim \sum_{n=1}^{\infty}\frac{1}{n}$ and therefore it diverges.
I don't understand the meaning of $\sim$ and the hole logic behind this answer. To me this doesn't look completly rigorous. Was here any of the convergence/divergence test implictly used?
 A: The symbol $\sim$ means asymptotically equivalent and we say that two sequences $(u_n)$ and $(v_n)$ are asymptotically equivalent and we write $u_n\sim v_n$ if 
$$u_n-v_n=o(u_n)\tag1$$
which's equivalent in the case $u_n\ne0$ for $n\ge n_0$ to
$$\lim_{n\to\infty}\frac{v_n}{u_n}=1$$
Now from $(1)$ we see that $\sum_n u_n$ is convergent if and only if $\sum_n v_n$ is convergent.
A: I completely agree with you. This argument does not look very rigorous, though it does provide some sort of intuition. (Edit: Note that @sami has provided a nice rigorous interpretation of the problem)
To provide a more rigorous approach, note that the limit comparison test works nicely here.
Let 
\begin{align*}
a_n &= \frac{1}{\sqrt{n^2+1}}\left(\frac{n}{n+1}\right)^n & b_n &= \frac{1}{n}
\end{align*}
Then 
$$
\lim_{n\to\infty}\frac{a_n}{b_n}=\frac{1}{e}
$$
(see if you can prove this!)
The limit comparison test then implies that either both $\sum a_n$ and $\sum b_n$ converge or both diverge. Of course $\sum b_n$ is harmonic so $\sum a_n$ diverges.
A: The $∼$ symbol  means that both expresions have similar behavior in infinity,
For instance consider the series $\sum a_n$ , $\sum b_n$
Where 
$a_n=\frac{1}{n}$ and  $b_n=\frac{1}{n}\frac{1}{e}$ 
So by Limit comparison test,
$\displaystyle{\lim_{x\to \infty} \frac{a_n}{b_n}=e>0}$
As we know armonic series diverges therefore $\sum _{n=1}^{\infty }\frac{1}{n}\frac{1}{e}$ diverges.
This could be helpful:
http://en.wikipedia.org/wiki/Limit_comparison_test
A: There's one more way. Consider 
$$
a_n = \frac{1}{\sqrt{n^2+1}} (\frac{n}{n+1} )^n = e^{\log \frac{1}{\sqrt{n^2+1}} (\frac{n}{n+1} )^n} = e^{-\frac{1}{2}\log (n^2+1) +n \log \frac{n}{n+1}}
$$
Here it's useful to use the Taylor expansion for the logarithm function. Asymptotically as $x \to 0, \ \log (1+x) \sim x$ and the property of logarithm: $\log a^t = t \log a$. Therefore,
$$
a_n  = e^{-\log n} \cdot e^{-\frac{1}{2n^2}} \cdot e^{-\frac{n}{n+1}} \geq e^{-\log n } e^{-1} e^{-1} = \frac{e^{-2}}{n} =b_n
$$
which, just like you have shown, diverges because $b_n \sim \frac{1}{n}$, which is known as Harmonic series. 
A: Squeezing if for the ultimate non-believers. Since $2\leq\left(1+\frac{1}{n}\right)^n\leq e$ and $n\leq\sqrt{n^2+1}\leq(n+1)$,
$$\sum_{n=1}^{N}\frac{1}{e(n+1)}\leq\sum_{n=1}^{N}a_n \leq \sum_{n=1}^{N}\frac{1}{2n},$$
but the LHS is greater than:
$$\frac{1}{e}\sum_{n=1}^{N}\log\left(1+\frac{1}{n+1}\right) = \frac{1}{e}\log\frac{N+2}{2}.$$
