Does this infinite series converge or diverge and why? 
$$
\sum_{k = 1}^\infty\sin\left(\frac1k + k\pi\right)
$$

I was thinking of using alternating series but I am not sure how to prove that is is alternating or decreasing.
 A: $$\sum_{k=1}^\infty (-1)^k\sin\Big(\frac{1}{k}\Big) = \sum_{k=1}^\infty x_k$$
with $x_k = (-1)^k\sin\Big(\frac{1}{k}\Big)$
Because $\sin(x)$ is increasing when $0 \leq x \leq \frac{\pi}{2}$ ;
$$\frac{1}{k+1} \leq \frac{1}{k} \Rightarrow \sin(\frac{1}{k+1}) \leq \sin(\frac{1}{k})$$
So we have $$|x_{k+1}| \leq |x_k|$$
We have too $$\lim\limits_{k\to\infty}(-1)^k\sin\Big(\frac{1}{k}\Big) = 0$$
$$\text{We conclude that }\sum_{k=1}^\infty x_k\text{ converges.}$$
A: The given series is $\sum_{k\geq 1}(-1)^{k}\sin\tfrac{1}{k}$ which is conditionally convergent by Leibniz' test, sic et simpliciter. By the inverse Laplace transform, such series equals
$$ \int_{0}^{+\infty}\sum_{k\geq 1}(-1)^k \sum_{n\geq 0}\frac{(-1)^n x^{2n}}{(2n)!(2n+1)!}e^{-kx}\,dx = -\int_{0}^{+\infty}\frac{\text{Ke}(x)}{e^x+1}\,dx $$
where $\text{Ke}$ is related to Kelvin and Bessel functions. An alternative representation is given by
$$ \sum_{n\geq 0}\sum_{k\geq 1}\frac{(-1)^n(-1)^{k+1}}{(2n+1)! k^{2n+1}}=-\sum_{n\geq 0}\frac{(1-4^{-n})(-1)^n\,\zeta(2n+1)}{(2n+1)!} $$
which is equally well suited for numerical purposes.
