Explain why series divergent or convergent See question title.
The series is as follows:
$$\sum_{n=1}^\infty{2n+\sin{n}\over {e^n-\cos n}}$$
Now, common sense dictates that the numerator "goes to infinity" much slower than the denominator, therefore the series is convergent. How to show this in more formal terms?
Thanks in advance for any and all advice.
 A: We have $$\sum_{n\geq1}\frac{2n+\sin\left(n\right)}{e^{n}-\cos\left(n\right)}\leq\sum_{n\geq1}\frac{2n+1}{e^{n}-1}\sim\sum_{n\geq1}\frac{2n}{e^{n}}=\frac{2e}{\left(e-1\right)^{2}}.$$
A: One way is to compare the terms to
$$
\frac{1}{2^n}
$$
or something else that is clearly larger (for $n\geq10$, at least) and just as clearly converges.
A: Use \begin{align}\lim_{n\to\infty}\frac{a_{n+1}}{a_n}&=\lim_{n\to\infty}\frac{{2(n+1)+\sin{(n+1)}\over {e^{n+1}-\cos (n+1)}}}{2n+\sin{n}\over {e^n-\cos n}}\\
&=\lim_{n\to\infty}\frac{2(n+1)+\sin{(n+1)}}{2n+\sin{n}}\times\frac{e^{n}-\cos n}{e^{n+1}-\cos (n+1)}\\
&=\lim_{n\to\infty}\frac{2n(1+\frac{1}{n}+\frac{\sin{(n+1)}}{2n})}{2n(1+\frac{\sin{n}}{2n})}\times\frac{e^{n}(1-\frac{\cos n}{e^n})}{e^{n+1}(1-\frac{\cos (n+1)}{e^{n+1}})}\\
&=\lim_{n\to\infty}\frac{(1+\frac{1}{n}+\frac{\sin{(n+1)}}{2n})}{(1+\frac{\sin{n}}{2n})}\times\frac{(1-\frac{\cos n}{e^n})}{e(1-\frac{\cos (n+1)}{e^{n+1}})}\\
&=\color{red}{\frac{(1+0+0)}{(1+0)}\times\frac{(1-0)}{e(1-0)}}\\
&=\frac{1}{e}
\end{align}
Note that since $-1\leq \sin x\leq 1, -1\leq \cos x\leq 1\Rightarrow \lim_{n\to\infty}\frac{\cos n}{e^{n}}=0~\text{and}~\lim_{n\to\infty}\frac{\sin n}{n}=0$
