0
$\begingroup$

I had three problems to work on and I was able to solve the third summation problem. The first two, I am having difficulty understanding as to how to proceed.

Here are the questions:

$$\sum_{n=1}^\infty \frac{n}{4^n};.$$

After using the series:

$$\sum_{n=0}^\infty {n x^n = \frac{x}{(1-x)^2}};.$$

I get

$$ \frac{4}{9} $$

Which is similar to result from Wolfram

I am not sure how to proceed on this one either:

$$\sum_{n=1}^\infty \frac{1}{n^2 - 4};.$$

I was able to solve the third one

$$\sum_{n=1}^\infty \frac{ln(n)}{n^3};.$$

For this problem, I referred to : Series simplification

for help and it helped me understand what steps I needed to do solve it.

Any guidance and help would be highly appreciated. Thanks! -SG

$\endgroup$
2
1
$\begingroup$

For the first one,

see this to find $$\sum_{k=1}^\infty[a+(k-1)d]r^{k-1}=\frac a{1-r}+\frac{rd}{(1-r)^2}$$

Can you recognize $a,d,k,r$ here?

For the second,

$$\frac4{n^2-4}=\frac{n+2-(n-2)}{(n+2)(n-2)}=\frac1{n-2}-\frac1{n+2}$$

Set a few values of $n$

$\endgroup$
5
  • $\begingroup$ So for the first one: $\endgroup$ – user40929 Feb 18 '15 at 7:07
  • $\begingroup$ @user40929, Set $a=k=d=1; r=1/4$ $\endgroup$ – lab bhattacharjee Feb 18 '15 at 7:08
  • $\begingroup$ I get 4/9 when I use the formula mentioned in my question and it coincides with the Wolfram Answer, but I get a different answer when I substitute the values in your series. Am I doing something wrong? $\endgroup$ – user40929 Feb 18 '15 at 7:28
  • $\begingroup$ I am getting an indefinite answer for the second one when I expand the series. Would that match your calculation? $\endgroup$ – user40929 Feb 18 '15 at 8:04
  • $\begingroup$ @user40929, Observe that the survivor terms are $$\frac1{-1}+\frac10+\frac12+\frac13$$. I think the sum should start with $n=3$ $\endgroup$ – lab bhattacharjee Feb 18 '15 at 11:18

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.