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?

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:
$$\frac{1}{e}\sum_{n=1}^{N}\log\left(1+\frac{1}{n+1}\right) = \frac{1}{e}\log\frac{N+2}{2}.$$
