<s>[INST] How to show that this series converges? I found in a book the following exercise:  

Show that the series $$\sum_{n=1}^{+\infty} \sin\left(\frac{n^2+n+1}{n+1}\right)$$ converges.  

At first sight, I tried to break the fraction and rewrite the series as $$\sum_{n=1}^{+\infty} \sin\left(n+\frac{1}{n+1}\right)$$
This actually behaves (almost) like $$\sum_{n=1}^{+\infty} \sin(n)+C$$ where $$C=\sum_{n=1}^{+\infty} \frac{\cos(n)}{n+1}$$
How can this series converge? Am I missing something here?
 [/INST] Suppose that
$$\lim_{n\to\infty}\sin\left(\frac{n^2+n+1}{n+1}\right)=\lim_{n\to\infty}\sin\left(n+\frac{1}{n+1}\right)=0$$
We have
$$\sin\left(n+1+\tfrac{1}{n+2}\right)=\sin\left(n+\tfrac{1}{n+1}\right)\cos\left(1+\tfrac{1}{n+2}-\tfrac{1}{n+1}\right)+\cos\left(n+\tfrac{1}{n+1}\right)\sin\left(1+\tfrac{1}{n+2}-\tfrac{1}{n+1}\right) $$
Since
$$\lim_{n\to\infty}\sin\left(n+\tfrac{1}{n+1}\right)=0 \\
\cos\left(1+\tfrac{1}{n+2}-\tfrac{1}{n+1}\right)\to\cos(1)\\
\sin\left(1+\tfrac{1}{n+2}-\tfrac{1}{n+1}\right)\to\sin(1)\ne 0$$
it follows that
$$\lim_{n\to\infty}\cos\left(n+\frac{1}{n+1}\right)=0 $$
However,
$$\sin^2\left(n+\frac{1}{n+1}\right)+\cos^2\left(n+\frac{1}{n+1}\right)=1 $$
and we get a contradiction. Therefore, the limit of the summand is not $0$, and the series diverges.
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