# 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., $L=1$).

• For the first one do you the limit of the coefficients as $k\to\infty$? (hint: Look at the definition of hyperbolic sine) Is it equal to zero or not? For the second, try the root test. – a... Dec 1 '14 at 16:13
• Hint for the second one: $(1-1/k)^k$ goes to $1/e$, so $(1-1/k)^{k^2}$ is, as $k$ grows, close to $1/e^{k}$. – Arthur Dec 1 '14 at 16:21
• Too late to edit the comment. I had meant to say do you know the limit of the terms. Btw for the second one recall a limit involving the exponential function: $\exp(x)=\lim\limits_{k\to\infty}\left(1+x/k\right)^k$ – a... Dec 1 '14 at 16:23

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.

• The idea of using the root test is right but the conclusion is wrong $$\limsup\limits_{n\to\infty}|a_n|^{\frac{1}{n}}<1\implies \sum\limits_{n=1}^\infty a_n\,\text{converges absolutely}$$ – a... Dec 1 '14 at 19:35
• but I have not used $|u_n|$ its only $u_n$ – Learnmore Dec 25 '14 at 3:20
• why so many down votes @BrianBóruma – Learnmore Dec 25 '14 at 3:26
• @BrianBóruma Unless this was edited, the user has given a correct answer. – Pedro Tamaroff Dec 25 '14 at 3:38
• ^The original version said the series diverges instead of converges. It wasn't edited until today. Also, a technical note: the answer should say $\lim_{n \to \infty}(1-\tfrac{1}{n})^n = e^{-1}$ instead of $(1-\tfrac{1}{n})^n = e^{-1}$, which is false for any integer $n$. – JimmyK4542 Dec 25 '14 at 3:41

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.