Series simplification

Question

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

Answer 1

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 $n$