Evaluate $\sum_{n=1}^{\infty} \frac{2n+1}{(n^{2} +n)^{2}}$ 
Evaluate  $$\sum_{n=1}^{\infty} \frac{2n+1}{(n^{2}+n)^{2}}.$$

I am getting two different results by using two different methods -
First Method
The above sum can be written as 
\begin{align}\sum_{n=1}^{N} (1/n^{2}  - 1/(n+1)^{2})&= 1 - 1/4 + 1/4 - 1/9 \dots -1/(N+1)^{2}\\ &= 1 - 1/(N+1)^{2} \end{align}
Taking the limit as $N\to\infty$, we have the the sum equal to $1$.
Second Method
Above sum is equal to -
\begin{align}\int_{1}^{\infty} \frac{2x+1}{(x^{2}+x)^{2}}\,dx\end{align}
Put $x^2 + x = t$
$$\int_{2}^{\infty} dt/t^{2}$$
$=[-1/t]_{2}^{\infty}$
$=1/2$
Why are these methods are giving different results?

In context, hopefully not too trivial.
Let $f(n)=\dfrac{2n+1}{(n^2+n)^2}$, $f(n)$ is strictly decreasing.
1)Your sum $\sum_{1}^{\infty}f(n)$ is an upper sum
for the integral  $\int_{1}^{\infty}f(x)dx$.
$U :=\sum_{1}^{\infty}f(n)=1$;
2)Now consider the lower sum:
$\sum_{2}^{\infty}f(n)$ for the integral.
$L := \sum_{2}^{\infty}f(n)=1/4;$
We have
$L =1/4 < 1/2$ (Integral)$ < U=1$.
See:
https://en.m.wikipedia.org/wiki/Integral_test_for_convergence
Link given by Zipirovic.
