Convergence of $\sum\limits_{n=1}^\infty \sin^2(\pi(n + \frac{1}{n})) $? My friend was practicing for his entrance examination and one of the problems on older exams was this one. It asks if it converges and to explain the reasoning behind that answer. The series is:
$$\sum_{n=1}^\infty \sin^2\left(\pi(n + \tfrac{1}{n})\right)$$
We had a few ideas, but none of them seemed to be working. Hoping for some help!
 A: Using $\lim_{x \to 0} \frac{\sin x}{x} = 1$ we can write
$$\sin^2 \left(\pi \left( n+\frac{1}{n} \right) \right) = \sin^2 \frac{\pi}{n} \sim \frac{\pi^2}{n^2}, \quad n \to \infty.$$
Here $f \sim g$ means $\frac{f}{g} \to 1$ (and thus is bounded from both sides). Since the series $\sum \frac{1}{n^2}$ converges (actually $= \frac{\pi^2}{6}$) by the comparison test your series also converges.
A: We have
$$\sin^2\left(\pi\left(n+\frac 1 n\right)\right)=\sin^2\left(\frac{\pi}{n}\right)=\frac{\pi^2}{n^2}+o\left(\frac{1}{n^2}\right)$$
so we conclude that the given series is convergent by comparison with the Riemann series.
A: *

*For every $x$ in $[0,\pi]$, $0\leqslant\sin(x)\leqslant x$ hence $\sin^2(x)\leqslant x^2$. 

*For every $n\geqslant1$, $\sin^2(\pi(n+1/n))=\sin^2(\pi/n)$ and $\pi/n$ is in $[0,\pi]$ hence $\sin^2(\pi/n)\leqslant\pi^2/n^2$. 

*Finally, the (Riemann) series $\sum\limits_{n\geqslant1}1/n^2$ converges hence, by comparison of series with nonnegative terms, the series $\sum\limits_{n\geqslant1}\sin^2(\pi(n+1/n))$ converges (and its sum is at most $\pi^4/6$).
