<s>[INST] prove that $\sum_{n=1}^\infty \frac{2^n+n^2+n}{2^{n+1}n(n+1)}$ is convergent and find the limit when $n \to \infty$ Does the following sumatory converges? If yes find the limit when $n \to \infty$ $$\sum_{n=1}^\infty \frac{2^n+n^2+n}{2^{n+1}n(n+1)}$$
ideas?
I have tried by the comparison test.
 [/INST] Observe that we can break up the sum to:
$$ \sum_{n=1}^{\infty} \frac{2^n}{2^{n+1} n(n+1)} + \frac{n^2+n}{2^{n+1}n(n+1)} $$
Which then becomes two sums:
$$ \sum_{n=1}^{\infty} \frac{1}{2n(n+1)} + \sum_{n=1}^{\infty}\frac{1}{2^{n+1}} $$
Now observe the left sum decays quadratically (any function f(n) which has the property that $\lim_{n\rightarrow \infty} f(n)/n = \infty$  similarly has the property that $\sum \frac{1}{f(n)}$ converges). 
Likewise the right hand sum is an easily recognized geometric sum. So now we can proceed to conclude the sum of the sums, itself must converge, and now the burden is of finding a closed form.
RIGHT SUM:
This is much easier so we warm up with it:
$$ \sum_{n=1}^{\infty} \frac{1}{2^{n+1}} = \frac{1}{2} \sum_{n=1}^{\infty} \frac{1}{2^n}  = \frac{1}{2} \left( \frac{1}{1 - \frac{1}{2}} - 1 \right) = \frac{1}{2}$$
LEFT SUM:
We can factor out the 2 to yield:
$$ \frac{1}{2} \sum_{n=1}^{\infty} \frac{1}{n(n+1)}$$
Which is 
$$ \frac{1}{2} \sum_{n=1}^{\infty} \left( \frac{1}{n} - \frac{1}{n+1} \right)$$
So now evluating this sum: is evaluation
$$ 1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3}  - \frac{1}{4} ... $$
Which at $n^{th}$ sum is $1  - \frac{1}{n+1}$ so as $n \rightarrow \infty$ this is just 1. Meaning:
$$ \frac{1}{2} \sum_{n=1}^{\infty} \frac{1}{n(n+1)} = \frac{1}{2}$$
thus:
$$\sum_{n=1}^{\infty} \frac{1}{2n(n+1)} + \sum_{n=1}^{\infty}\frac{1}{2^{n+1}} = \frac{1}{2} + \frac{1}{2} = 1$$
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