Consider the following series $$\sum_{n \geq 1} \sin \left(\frac{n^2+n+1}{n+1} \pi\right)$$

The general term of the series does not go to zero, in fact $$\nexists\lim_{n \to \infty} \sin \left(\frac{n^2+n+1}{n+1} \pi\right) $$

Nevertheless on textbook I find that $$\sum_{n \geq 1} \sin \left(\frac{n^2+n+1}{n+1} \pi\right) = \sum_{n \geq 1} (-1)^n \sin \left(\frac{\pi}{n+1} \right)$$ Which converges conditionally.

I understand how to get the last series and why it converges conditionally, but I always thought that a necessary condition for any convergence of a series is that the limit of the general term is zero.

Am I missing something?

  • $\begingroup$ Yes, you are missing that the limit of the general term is zero. You simply assert it doesn't converge. $\endgroup$ – Thomas Andrews Nov 12 '16 at 14:26
  • $\begingroup$ The sum will fail to be cauchy. On the other hand, integrals like $\int_0^\infty\sin(x^2)dx$ can converge. $\endgroup$ – Simply Beautiful Art Nov 12 '16 at 14:41
  • $\begingroup$ @SimpleArt The series does converge. $\endgroup$ – Mark Viola Nov 12 '16 at 14:47
  • 1
    $\begingroup$ @SimpleArt: the series converges and is therefore Cauchy. $\endgroup$ – robjohn Nov 12 '16 at 14:48
  • $\begingroup$ @Dr.MV I mean if the $n$th term didn't approach $0$. Sorry, should've been more clear on that. $\endgroup$ – Simply Beautiful Art Nov 12 '16 at 14:49

Note that $$\frac{n^2+n+1}{n+1}=n+\frac{1}{n+1}$$

Therefore, we can write

$$\sin\left(\frac{n^2+n+1}{n+1}\,\pi\right)=\sin\left(n\pi +\frac{\pi}{n+1}\right)=(-1)^n\sin\left(\frac{\pi}{n+1}\right)$$

Certainly, we have $\lim_{n\to \infty}(-1)^n\sin\left(\frac{\pi}{n+1}\right)=0$ and hence $\lim_{n\to \infty}\sin\left(\frac{n^2+n+1}{n+1}\,\pi\right)=0$ also.

So, the general terms of the series do approach $0$. And in fact, by Leibniz's test for alternating series, we assert that

$$\sum_{n=1}^\infty \sin\left(\frac{n^2+n+1}{n+1}\,\pi\right)=\sum_{n=1}^\infty (-1)^n\sin\left(\frac{\pi}{n+1}\right)$$



This is a roundabout demonstration of convergence, but the main point is the acceleration of convergence.

The sum converges pretty slowly, but we can accelerate the convergence $$ \begin{align} \sum_{n=1}^\infty(-1)^n\sin\left(\frac\pi{n+1}\right) &=\sum_{n=1}^\infty(-1)^n\sum_{k=0}^\infty(-1)^k\frac{\pi^{2k+1}}{(2k+1)!(n+1)^{2k+1}}\\ &=\sum_{n=1}^\infty(-1)^n\frac\pi{n+1}+\sum_{k=1}^\infty(-1)^k\frac{\pi^{2k+1}}{(2k+1)!}\left(\tfrac{4^k-1}{4^k}\zeta(2k+1)-1\right)\\ &=\pi(\log(2)\color{#C00000}{-1})+\sum_{k=1}^\infty(-1)^k\frac{\pi^{2k+1}}{(2k+1)!}\left(\tfrac{4^k-1}{4^k}\zeta(2k+1)\color{#C00000}{-1}\right)\\[6pt] &=-0.52202133938400117822 \end{align} $$ Note that the sum of the terms in red is $-\sin(\pi)=0$, but the approximation $$ \tfrac{4^k-1}{4^k}\zeta(2k+1)-1\sim-\frac1{2^{2k+1}} $$ shows that they accelerate the convergence.


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.