<s>[INST] Calculus - Prove that $\{a_n\}$ converges I need to prove that $a_n$ converges :
$$a_n=\frac{1}{2+1}+\frac{1}{2^2+2}+\cdots+\frac{1}{2^n+n}$$ 
My solution
Let choose $\epsilon=1$, Exist $N>0$ so all $N \leq x$ appiles $|a_n-L|<\epsilon$
Therefore :
$$\left|\frac{1}{2^n+n}-L\right|<\epsilon=1$$
$$L-1<\frac{1}{2^n+n}<L+1$$
Therefore the function is bounded.
Also $a_n$ is increasing since $a_{n+1}-a_n>0$
$$a_n = \frac{1}{2^n+n}$$
$$a_{n+1} = \frac{1}{2^{(n+1)}+n+1}$$
$$a_{n+1}-a_n=\frac{1}{2^n+n}-\frac{1}{2^{(n+1)}+n+1}=\frac{2^{n+1}+n+1-2^n-n}{(2^n+n)(2^{n+1}+n+1)}=\frac{2^{n+1}+1-2^n}{(2^n+n)(2^{n+1}+n+1)}>0 (1 \leq n)$$
As well it clear that the function is continuous since $1,2^n,n$ are continuous.
Therefore it possible to conclude that the function is converges.
Is this proof perfect?
Thanks!
 [/INST] Hint:
Notice that $$\frac{1}{2^n+n} < \frac{1}{2^n}$$
And use the Comparison Test. 
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