proof that $\sum_{n=1}^{\infty}\frac{n}{n^2+2n+1}$ diverges, by comparsion I need to prove that 
$$\sum_{n=1}^{\infty}\frac{n}{n^2+2n+1}$$ 
diverges by comparsion.
The way I did was to use 
$$\frac{n}{n^2+2n+1}>\frac{n}{n^2+2n^2+n^2} = \frac{n}{4n^2} = \frac{1}{4n}$$
which diverges. Can I do that? Because $$2n+1<2n^2+n^2\implies n^2+2n+1<n^2+2n^2+n^2\implies $$
$$\frac{1}{n^2+2n+1}>\frac{1}{n^2+2n^2+n^2}$$
for large $n$.
I also tried seeing 
$$\sum_{n=1}^{\infty}\frac{n}{n^2+2n+1} = \sum_{n=1}^{\infty}\frac{n}{(n+1)^2}$$ 
but I couldn't find any comparsion for that. Am I right in the first demonstration? Can you find a comparsion for the second attempt I did?

$$\frac{n}{n^2+2n+1} = \frac{n}{(n+1)^2} = \frac{n + 1 - 1}{(n+1)^2} = \frac{1}{(n+1)} - \frac{1}{(n+1)^2}$$
Which diverges because $\frac{1}{n+1}$ does so and $\frac{-1}{(n+1)^2}$ doesn't.
