Why does $\sum_{n=0}^\infty (\sin{\frac{1}{n}})(\sinh{\frac{1}{n}})(\cos n)$ converge $$\sum_{n=1}^\infty (\sin{\frac{1}{n}})(\sinh{\frac{1}{n}})(\cos n)$$
I tried this
$(\sin{\frac{1}{n}})(\sinh{\frac{1}{n}})(\cos n)\le \sinh{\frac{1}{n}}$ and then i want to prove that $\sinh{\frac{1}{n}}$ converges. I am kind of stuck at this point.
 A: Because 
$$\lim_{x\rightarrow 0} \frac{\sin(x)}{x}=\lim_{x\rightarrow 0} \frac{\sinh(x)}{x}=1$$
thus 
$$\lvert \sin \left( \frac{1}{n} \right)\sinh \left( \frac{1}{n} \right)\cos(n)\rvert \leq \lvert \sin \left( \frac{1}{n} \right) \sinh \left( \frac{1}{n} \right) \rvert \simeq \frac{1}{n^2}$$
Prove that $\lim\limits_{x \to 0} \sinh(x)/x =1$.
A: You can partition the set $\mathbb{N}\bigcap[1,\infty)$ into a countable set of sets $\{A_1, A_2, ... \}$, where the cardinality of $A_i$ is less than 5, and $A_i$ contains consecutive integers only, and such that $n_1,n_2\in A_i \Rightarrow\text{sign}(\cos(n_1)) =\text{sign}(\cos(n_2))$.
Then because of the finite cardinality and convergence towards zero of the terms in your sum
$$\lim_{i\to\infty}\sum_{n\in A_i}(\sin{\frac{1}{n}})(\sinh{\frac{1}{n}})\cos(n)=0$$
The following two sums are the same because of the consecutiveness in the $A_i$ sets, and the first is alternating and has terms converging to zero which makes it convergent:
$$\infty>\sum_{i=1}^{\infty}\sum_{n\in A_i}(\sin{\frac{1}{n}})(\sinh{\frac{1}{n}})\cos(n)=\sum_{n=1}^{\infty}(\sin{\frac{1}{n}})(\sinh{\frac{1}{n}})\cos(n)$$
A: It converges absolutely because
$$\biggl\lvert\sin\frac1n\cdot\sinh\frac1n\cdot\cos n\biggr\rvert=_\infty O\biggl(\frac 1{n^2}\biggr),$$
and the series  $\displaystyle\sum_{n\ge 1}\frac1{n^2}\;$ converges.
