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$$ \sum_{k = 1}^\infty\sin\left(\frac1k + k\pi\right) $$

I was thinking of using alternating series but I am not sure how to prove that is is alternating or decreasing.

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    $\begingroup$ it's useful to observe $\sin(\frac{1}{k}+k\pi)=(-1)^k\sin(\frac{1}{k})$. Do you know Leibniz criterion? $\endgroup$ – Lucio Oct 31 '17 at 11:35
  • $\begingroup$ It kinda feels like it converges since the sequence tends to sin(0+kπ)≈0 and is alternating, so for big k you are adding a number and then substracting the "same" number. I'm on my phone right now so I'd love to help some more but can't. I'd need paper $\endgroup$ – Francisco José Letterio Oct 31 '17 at 12:04
  • $\begingroup$ Lucio your comment was just the hint I needed. You need to expand the sine expression to get it in the form of an alternating series. Thank you. $\endgroup$ – BuluBestTapu Oct 31 '17 at 12:07
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$$\sum_{k=1}^\infty (-1)^k\sin\Big(\frac{1}{k}\Big) = \sum_{k=1}^\infty x_k$$ 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 $$\lim\limits_{k\to\infty}(-1)^k\sin\Big(\frac{1}{k}\Big) = 0$$

$$\text{We conclude that }\sum_{k=1}^\infty x_k\text{ converges.}$$

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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.

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