Convergence/Divergence of series with terms $2^{n}\left ( \frac{n}{n+1} \right )^{n^{2}} $ and $\sin (n)\sin \frac{x}{n}$ Help me please with these 2 questions:
1.Does it converge or diverge? :
$$ \sum_{n=2}^{\infty }2^{n}\left ( \frac{n}{n+1} \right )^{n^{2}} $$
2.Check out absolute and conditional convergence of:  $x>0 $
$$ \sum_{n=1}^{\infty }\sin (n)\sin \frac{x}{n} $$
Thanks a lot!
 A: Hint for 1:
For sufficiently large $n$, $(\frac{n}{n+1})^n = (1 - \frac{1}{n+1})^n \le c$ for some $ 0 \lt c \lt \frac{1}{2}$. Why?
Now trying using the above to prove that your series converges.
For part 2, I believe you can use the Dirichlet Test to prove convergence.
To show that the series does not converge absolutely, use $\sin (x/n) \ge x/2n$ for sufficiently large $n$ and use the fact that at least one of $n$, $n+1$ is more than $\frac{1}{2}$ away from the multiple of $\pi$ which is closest to them.
A: Part 2
As indicated in Aryabatha's answer, convergence of $\sum(\sin n)\sin(x/n)$ follows from Dirichlet's test: $\sin(x/n)$ is eventually decreasing, converges to $0$ and the partial sums $\sum_{k=1}^n\sin k$ are bounded. To show that it does not converge absolutely, use the inequalities
$$
\sin x\ge \frac{2\,x}\pi,\quad|\sin n|\ge\sin^2n=\frac{1-\cos(2\,n)}{2}.
$$
Then
$$
|(\sin n)\sin\Bigl(\frac{x}{n}\Bigr)|\ge\frac{x}{\pi}\Bigl(\frac1n-\frac{\cos(2\,n)}{n}\Bigr).
$$
Again by Dirichlet'e test $\sum_{n=1}^\infty\cos(2\,n)/n$ converges. In particular there exists a constant $A>0$ such that $\Bigl|\sum_{k=1}^n\cos(2\,n)/n\Bigr|\le A$ for all $n$. Then
$$
\sum_{k=1}^n|(\sin k)\sin\Bigl(\dfrac{x}{k}\Bigr)|\ge\frac{x}{\pi}\Bigl(\sum_{k=1}^n\frac1k-\sum_{k=1}^n\frac{\cos(2\,n)}{n}\Bigr)\ge\frac{x}{\pi}(\log(n+1)-A).
$$
