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? 
 A: 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.
A: 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.
