Q: For which of the following choice of $a_k$ is $\sum a_k$ convergent? For which of the following choice of $a_k$ is $\sum a_k$ convergent?
i)$\displaystyle \frac {\sinh(k)}{2^k}$
ii)$\displaystyle \bigg(1-\frac{1}{k}\bigg)^{k^2}$

Honestly, I have no idea. Usually, when I see $\sin$ or $\cos$, I use consider absolute convergence, since it is easier; however, clearly this will not work since $\displaystyle \sum\frac{1}{k}$ diverges. 
I considered using the integral test, but am not sure actually how to properly use it. 
For the second, was switching between partial sums and Comparison Test. I tried Ratio Test, but it didn't produce a result (i.e., $L=1$).

A: For the first one:
\begin{align}
  \lim_{k\to\infty}\frac{\sinh(k)}{2^k}&=\lim_{k\to\infty}\frac{e^k - e^{-k}}{2^{k+1}} \\
&= \lim_{k\to\infty}\frac{e^k}{2^{k+1}} \to\infty
\end{align}
Since the $k$th term does not go to zero as $k\to\infty$, we know the first diverges.

A: For the 2nd one consider Cauchy's root test
$$u_n^{\frac{1}{n}}=\left(1-\dfrac{1}{n}\right)^n\longrightarrow e^{-1}<1$$
Hence $\sum (1-k^{-1})^{k^2}$ converges. 
