Question

Find the sum of $n$ terms of the series whose $n^{th}$ term is $\frac{n^4+n^2+1}{n^4+n}$.

Answer 1

$T_n=\frac{n^4+n^2+1}{n^4+n}=1+\frac{n^2-n+1}{n(n^3+1)}=1+\frac{1}{n(n+1)}=1+\frac{1}{n}-\frac{1}{n+1}$

$S_n=\sum T_n=n+(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\dots +\frac{1}{n}-\frac{1}{n+1})$

$S_n=n+1-\frac{1}{n+1}$