# 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?

• It cannot converge, right? If $\sum_n a_n$ converges then $a_n \to 0$, which does not happen here. – user58955 Feb 12 '20 at 9:21
• @user58955 so, the exercise is wrong obviously ? – Konstantinos Gaitanas Feb 12 '20 at 9:24
• 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 $[-1, 1]$... – 5xum Feb 12 '20 at 9:24
• @KonstantinosGaitanas from what book this exercise comes from? – Masacroso Feb 12 '20 at 11:50
• @Masacroso from "Analysis I" by prof. Pantelidis. It was given to me when I was undergraduate and I gave it a look yesterday. – Konstantinos Gaitanas Feb 12 '20 at 12:12

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.