<s>[INST] Sums $a_k=[\frac{2+(-1)^k}{3^k}, (\frac{1}{n}-\frac{1}{n+2}), \frac{1}{4k^2-1},\sum_{l=0}^k{k \choose l}\frac{1}{2^{k+l}}]$ 
Determien the sums of the following series'.
1:$\sum_{k=0}^\infty \frac{2+(-1)^k}{3^k}$
2:$\sum_{k=0}^\infty (\frac{1}{n}-\frac{1}{n+2})$
3:$\sum_{k=0}^\infty \frac{1}{4k^2-1}$
4:$\sum_{k=0}^\infty \sum_{l=0}^k{k \choose l}\frac{1}{2^{k+l}}$

I have always been bad with infinite series', so I decided to practice them strongly for the next couple of days. I found an exercise in a textbook and kinda troubling with some of them.
Here's what I was thinking of doing.
1: I thought about dividing this into two sums, $\sum_{k=0}^\infty \frac{2}{3^k}+\sum_{k=0}^\infty \frac{(-1)^k}{3^k}$, which should be okay since they converge, right? Anyway, then I can see the geometric series there, giving me $3+\frac{3}{4}=3,75$. Is that correct?
2: I think that's what's called a telescope series? So I tried writing down some of the first numbers to see where it starts canceling each other out:
$\frac{1}{1}-\frac{1}{3}+\frac{1}{2}-\frac{1 }{ 4}+\frac{ 1}{ 3}-\frac{ 1}{ 5}+\frac{1 }{ 4}-\frac{1 }{6 } + ...$. Since $\frac{1}{n}$ converges to zero for $n\to \infty$, the sum of this infinite series is $\frac{3}{2}$?
About 3 and 4. I'm pretty much lost there. Can't seem to find an approach to find the sum.
Any tips?
 [/INST] $1-$
$$\sum_{k=0}^{\infty }\frac{2+(-1)^k}{3^k}=\sum_{k=0}^{\infty }\frac{2}{3^k}+\frac{(-1)^k}{3^n}=2\frac{1}{1-1/3}+\frac{1}{1+1/3}$$
$3-$
$$\sum_{k=0}^{\infty }\frac{1}{4k^2-1}=\sum_{k=0}^{\infty }0.5[\frac{1}{2k-1}-\frac{1}{2k+1}]=0.5[1/1-1/3+1/3-1/5+1/5+....]=\frac{1}{2}$$
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