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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?

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    $\begingroup$ It cannot converge, right? If $\sum_n a_n$ converges then $a_n \to 0$, which does not happen here. $\endgroup$ – user58955 Feb 12 '20 at 9:21
  • $\begingroup$ @user58955 so, the exercise is wrong obviously ? $\endgroup$ – Konstantinos Gaitanas Feb 12 '20 at 9:24
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    $\begingroup$ I would be very surprised if this series actually converged, as it would mean that $$\lim_{n\to\infty}\sin\left(n+\frac{1}{n+1}\right) = 0,$$ which would be very surprising, given that $\{\sin(n)|n\in\mathbb N\}$ is dense in $[-1, 1]$... $\endgroup$ – 5xum Feb 12 '20 at 9:24
  • $\begingroup$ @KonstantinosGaitanas from what book this exercise comes from? $\endgroup$ – Masacroso Feb 12 '20 at 11:50
  • $\begingroup$ @Masacroso from "Analysis I" by prof. Pantelidis. It was given to me when I was undergraduate and I gave it a look yesterday. $\endgroup$ – Konstantinos Gaitanas Feb 12 '20 at 12:12
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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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