<s>[INST] KVPY Question: Sequence and Series. 
$S_n = \cfrac{7}{4 \cdot 1 \cdot 2} + \cfrac{10}{4^2 \cdot 2 \cdot 3} + \cfrac{13}{4^3 \cdot 3 \cdot 4} + ....$, then $S_{\infty}=$

I found the general term:
$$\sum_{n=1}^{n \rightarrow \infty} \cfrac{(3n+4)}{n(n+1)4^{n}} = $$
How to evaluate further? Please help.

I am at an elementary level, so this question might seem dumb, sorry.
 [/INST] $$\frac{4n+3}{(n+1)4^n}=\frac 1{4^{n-1}\cdot n}-\frac 1{4^n\cdot (n+1)}$$
Now we give values $n=1 \rightarrow n$ and sum up and cancel similar terms :
$$\frac 1{4^0\cdot 1}-\frac 1{4\cdot 2}$$
$$\frac 1{4\cdot 2}-\frac 1{4^2\cdot 3}$$
$$\frac 1{4^2\cdot 3}-\frac 1{4^3\cdot 4}$$
$$\frac 1{4^3\cdot 4}-\frac 1{4^4\cdot 5}$$
$$\vdots$$
$$\frac 1{4^{n-2}\cdot(n-1) }-\frac 1{4^{n-1}\cdot n}$$
$$\frac 1{4^{n-1}\cdot n}-\frac 1{4^n\cdot (n+1)}$$
$$\therefore S_n=1-\frac 1{4^n(n+1)}$$
$$\Rightarrow \lim_{n\rightarrow \infty} S_n=1$$
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