Series simplification

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

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 $n$

• So for the first one: – user40929 Feb 18 '15 at 7:07
• @user40929, Set $a=k=d=1; r=1/4$ – lab bhattacharjee Feb 18 '15 at 7:08
• 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? – user40929 Feb 18 '15 at 7:28
• I am getting an indefinite answer for the second one when I expand the series. Would that match your calculation? – user40929 Feb 18 '15 at 8:04
• @user40929, Observe that the survivor terms are $$\frac1{-1}+\frac10+\frac12+\frac13$$. I think the sum should start with $n=3$ – lab bhattacharjee Feb 18 '15 at 11:18