# Can series converge even if the general term does not have limit?

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?

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

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

converges.

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.