I am asked to prove that $\sum\limits_{n=1}^\infty {\sin(n)\sin(n^2)\over n}$ converges using the following fact:
Let $(a_n)_{n=1}^\infty$ be a bounded sequence. Then $\sum\limits_{n=1}^\infty {a_{n+1}-a_n\over n}$ converges.
I think I have to solve the recurrence $a_{n+1} - a_n = \sin(n)\sin(n^2)$, but how? What are the inital conditions? What basis to look for the solution in?
Thanks.
The addition formula for cosines reads $\cos(a\pm b)=\cos(a)\cos(b)\mp\sin(a)\sin(b)$ and yields $$ 2\sin(n)\sin(n^2)=\cos(n^2-n)-\cos(n^2+n)=b_n-b_{n+1}, $$ where $b_n=\cos(n(n-1))$. Thus, for every $N\geqslant0$, $$ \sum_{n=0}^{N-1}\sin(n)\sin(n^2)=\frac{b_0-b_{N}}2=\frac{1-\cos(N(N-1))}2. $$ The RHS is always in $[0,1]$ and the LHS, using your notations, is $a_{N}-a_0$, hence your proof is complete.