Question

• Evaluate \(\displaystyle\sum_{n=1}^\infty\left(\frac{1}{n+n^2}\right)\)

Solution

Notice the partial fraction of \(\dfrac{1}{n+n^2}\) is

$$ \frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1} $$

Therefore

$$ \begin{align*} \sum_{n=1}^\infty\left(\frac{1}{n+n^2}\right)&=\sum_{n=1}^\infty\left(\frac{1}{n}-\frac{1}{n+1}\right)\\ &=\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\dots\\ &=1 \end{align*} $$