Finding the exact amount of a sigma problem? $$\begin{align*}
\sum_{k=1}^\infty \frac{1}{k(k+1)}
\end{align*}$$
This is a telescoping series; therefore I use partial fractions to solve.
$\int_{1}^{∞}  (1)/(k+1) $ = $ ((A/k)+(B)/(k+1))$
A= 1
B= -1
After solving for partial fractions I get.
$ (1/k)+(-1/k+1)$|∞
                 |1

Then I solve by subtracting
$(1/∞)-(1)/(∞+1)-(1/1+-1/1+1) = -1$ 
The answer is 1 yet I got -1. Some input on where I made a mistake would help.

You've made this more complicated than perhaps it needs to be. First note that $\displaystyle \frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1}$. Hence
$\displaystyle \sum_{k=1}^N \frac{1}{k(k+1)} = \sum_{k=1}^N \frac{1}{k} - \frac{1}{k+1} $
$\displaystyle  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \   = 1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - ... + \frac{1}{N} - \frac{1}{N+1} $
$\displaystyle  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 1 - \frac{1}{N+1}$
Now take the limit as $N \rightarrow \infty$.
