On the convergence of a trigonometric series If the following is duplicated, please let me know. 
Suppose that $a_n = \frac{\sin\frac 1n\sin\frac 1{n+1}}{\cos\frac 1{n(n+1)}}$. I wonder if the series $\sum a_n$ is convergent? And if so, then what is its sum?  Any suggestion would be helpful. 

Recall that the sine function satisfies the inequalities
$$x\cos x\le \sin x\le x \tag 1$$
for $0\le x\le \pi/2$.  From $(1)$, it is easy to show that fpr $0\le x\le \pi/2$, the cosine function satisfies the inequalities
$$\sqrt{1-x^2}\le \cos x\le 1 \tag 2$$
Then,  from $(1)$ and $(2)$, we can write 
$$0\le\sin\left(\frac1n\right)\sin\left(\frac{1}{n+1}\right)\le \frac{1}{n(n+1)}\tag 3$$
and 
$$0\le \frac{1}{\cos \left(\frac{1}{n(n+1)}\right)}\le \frac{1}{\sqrt{1-\left(\frac{1}{n(n+1)}\right)^2}} \tag 4$$
For $n\ge 1$, it is easy to arrive at the inequality
$$\frac{1}{\sqrt{1-\left(\frac{1}{n(n+1)}\right)^2}} \le 1+\frac1n \tag 5$$
Putting $(3)$, $(4)$, and $(5)$ together yields
$$0\le \frac{\sin\left(\frac1n\right)\sin\left(\frac{1}{n+1}\right)}{\cos \left(\frac{1}{n(n+1)}\right)}\le \frac{1}{n(n+1)\sqrt{1-\left(\frac{1}{n(n+1)}\right)^2}}\le \frac1{n^2}$$
Then, given that $\sum_{n=1}^\infty \frac1{n^2}=\frac{\pi^2}{6}$, we have
$$0\le \sum_{n=1}^\infty \frac{\sin\left(\frac1n\right)\sin\left(\frac{1}{n+1}\right)}{\cos \left(\frac{1}{n(n+1)}\right)}\le \frac{\pi^2}{6}$$
and we have not only proven convergence, but provided an upper bound.
