0
$\begingroup$

So I was working on this problem

$$\sum\limits_{n=2}^\infty \frac{2^n}{n^{100}}$$

After all of my work performing the ratio test, I ended up with a $$\frac{2}{n+1} = 0$$

Therefore, I assumed it absolutely converges, but turns out that it diverges. May I please get an explanation why?

$\endgroup$
3
$\begingroup$

I think you computed the ratio incorrectly. It should be $$\frac{2^{n+1}}{(n+1)^{100}} \cdot \frac{n^{100}}{2^n} = 2 \left(\frac{n}{n+1}\right)^{100}$$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ You're right, I did. I cancelled out the n in the numerator. Thanks for explaining. $\endgroup$ – Adan Vivero Apr 30 '19 at 23:07
4
$\begingroup$

A simple remark: you don't even need the ratio test, as this series diverges trivially: its general terù $\frac{2^n}{n^{100}}$ does not tend to $0$ at $\infty$.

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

$$\frac{2^{n+1}}{(n+1)^{100}}\frac{n^{100}}{2^n}=\frac{2}{(1+\frac{1}{n})^{100}}$$

| cite | improve this answer | |
$\endgroup$

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.