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., $L=1$).

  • $\begingroup$ 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. $\endgroup$ – a... Dec 1 '14 at 16:13
  • $\begingroup$ 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 $1/e^{k}$. $\endgroup$ – Arthur Dec 1 '14 at 16:21
  • $\begingroup$ 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: $\exp(x)=\lim\limits_{k\to\infty}\left(1+x/k\right)^k$ $\endgroup$ – 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.

  • 2
    $\begingroup$ The idea of using the root test is right but the conclusion is wrong $$\limsup\limits_{n\to\infty}|a_n|^{\frac{1}{n}}<1\implies \sum\limits_{n=1}^\infty a_n\,\text{converges absolutely}$$ $\endgroup$ – a... Dec 1 '14 at 19:35
  • $\begingroup$ but I have not used $|u_n| $ its only $u_n$ $\endgroup$ – Learnmore Dec 25 '14 at 3:20
  • $\begingroup$ why so many down votes @BrianBóruma $\endgroup$ – Learnmore Dec 25 '14 at 3:26
  • $\begingroup$ @BrianBóruma Unless this was edited, the user has given a correct answer. $\endgroup$ – Pedro Tamaroff Dec 25 '14 at 3:38
  • 4
    $\begingroup$ ^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 $n$. $\endgroup$ – 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.