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

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

  • 3
    $\begingroup$ 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

$$\sum_{k=1}^\infty (-1)^k\sin\Big(\frac{1}{k}\Big) = \sum_{k=1}^\infty x_k$$ 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 $$\lim\limits_{k\to\infty}(-1)^k\sin\Big(\frac{1}{k}\Big) = 0$$

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


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.