Why does this sum converge $\sum\limits_{k=1}^\infty\left (\frac{k\sin k}{2k+1}\right)^k$ I don't understand why this sum converges.
$$\sum\limits_{k=1}^\infty \left(\frac{k\sin k}{2k+1}\right)^k$$
$$\lim_{x\to\infty} \left(\frac{k\sin k}{2k+1}\right) = diverge$$
I don't find any other test to proof it, any idea?
Thanks!
 A: $$\sum\limits_{k=1}^\infty \left(\frac{k\sin k}{2k+1}\right)^k$$
First note that
$$\sum\limits_{k=1}^\infty\left|\frac{k\sin k}{2k+1}\right|^k\leq\sum\limits_{k=1}^\infty \left|\frac{k}{2k+1}\right|^k$$
Now lets use the root test on the series on the right hand side
$$r=\limsup\limits_{k\to\infty}\sqrt[k]{\left|\frac{k}{2k+1}\right|^k}=\limsup\limits_{k\to\infty}\left|\frac{k}{2k+1}\right|$$
$$=\limsup\limits_{k\to\infty}\left|\frac{1}{2+\frac1k}\right|=\frac12\lt 1$$
By the root test, we have
$$\sum\limits_{k=1}^\infty \left|\frac{k}{2k+1}\right|^k=\mbox{convergent}$$
Which implies that
$$\sum\limits_{k=1}^\infty \left(\frac{k}{2k+1}\right)^k=\mbox{absolutely convergent}$$
Therefore by the direct comparison test, we have
$$\sum\limits_{k=1}^\infty \left(\frac{k\sin k}{2k+1}\right)^k=\mbox{absolutely convergent}$$
And absolute convergence also implies convergence.
A: $\lim_{x \to \infty} \left( \frac{k \sin k}{2k+1} \right)$ does diverge, but it always remains in the interval $[-\frac{1}{2}, \frac{1}{2}]$. Therefore, as we raise it to increasing powers, we do indeed get something which tends to zero. That means it's plausible for the sum to converge.
Notice that 
$$\frac{k}{2k+1} \leq \frac{1}{2}$$
so the summand is less than or equal in modulus to $$\left \vert \left(\frac{\sin{k}}{2}\right)^k \right\vert$$
This converges absolutely by comparison with the geometric series with summand $2^{-k}$.
Therefore, by the comparison test, the series converges.
A: Note that $|\sin k|\leqslant 1$ for all $k$. So by the root test
$$\limsup_{k\to\infty}\frac{k\sin k}{2k+1} \leqslant \limsup_{k\to\infty} \frac k{2k+1} = \frac12<1, $$
and hence the series converges.
