Series, limits and convergence

Question

Can you help me with it with working

$\sum _{n=1}^{\infty }\:\cos \left(\frac{2n+1}{n^2+n}\right)\sin \left(\frac{1}{n^2+n}\right)$

I am struck in the part where i get,

$\lim _{n\to \infty \:\:}\left(n\left(\frac{\cos \left(\frac{2n+1}{n^2+n}\right)\sin \left(\frac{1}{n^2+n}\right)}{\cos \left(\frac{2\left(n+1\right)+1}{\left(n+1\right)^2+n+1}\right)\sin \left(\frac{1}{\left(n+1\right)^2+n+1}\right)}-1\right)\right)$

this expression

Answer 1

HINT

For the first one

$$\sum_{n=1}^\infty \frac{n+1}{(n+2)!}=\sum_{n=1}^\infty \frac{n+2-1}{(n+2)!}=\sum_{n=1}^\infty \frac{1}{(n+1)!}-\sum_{n=1}^\infty \frac{1}{(n+2)!}$$

then recall that

$$e=\sum_{n=0}^\infty \frac{1}{n!}$$

For the one presented after editing note that since $\frac{2n+1}{n^2+n}\to 0 $ and $\frac{1}{n^2+n}\to 0$ we have

$$\cos \left(\frac{2n+1}{n^2+n}\right)\sin \left(\frac{1}{n^2+n}\right)\sim1\cdot\frac{1}{n^2+n}$$

Answer 2

For large enough $n$ we have $$0<\cos \left({2n+1\over n^2+n}\right)\sin \left({1\over n^2+n}\right)<1\times {1\over n^2+n}={{1\over n^2+n}}$$therefore the series converges using Squeeze theorem.