Notice the partial fraction of 1n+n2\dfrac{1}{n+n^2}n+n21 is
1n(n+1)=1n−1n+1 \frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1} n(n+1)1=n1−n+11
Therefore
∑n=1∞(1n+n2)=∑n=1∞(1n−1n+1)=(1−12)+(12−13)+(13−14)+…=1 \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*} n=1∑∞(n+n21)=n=1∑∞(n1−n+11)=(1−21)+(21−31)+(31−41)+…=1
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