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?
 A: The Integral Test for series does NOT assert that the infinite series and the improper integral are equal to each other. In fact, from the proof, it can be seen that they can't possibly be equal to each other (except for maybe some carefully constructed step functions, if we relax the continuity requirement). The test only says that if the integral converges, then the series converges as well (and the same for divergence), but it does not provide a value for the series in this case (although it does provide a useful remainder estimate). In short:
$$\sum_{n=1}^{\infty}f(n)\color{red}{\neq}\int_1^{\infty}f(x)\,dx,$$
which is why your second "method" is wrong.
A: 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.
