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Question

Solution

Notice the partial fraction of 1n+n2\dfrac{1}{n+n^2} is

1n(n+1)=1n1n+1 \frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}

Therefore

n=1(1n+n2)=n=1(1n1n+1)=(112)+(1213)+(1314)+=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*}


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