The sum to $n$ terms of the series

$\frac{1}{1^3}+\frac{1+2}{1^3+2^3}+\frac{1+2+3}{1^3+2^3+3^3}+\dots$is

If $t_n$ denotes the nth term of the series, then

$\begin{array}{l}{t}_n=\frac{1+2+3+\cdots +n}{1^3+2^3+3^3+\cdots +n^3}=\frac{\frac{1}{2}n(n+1)}{\frac{1}{4}{n}^2(n+1{)}^2}\\ =\frac{2}{n(n+1)}=2\left[\frac{1}{n}-\frac{1}{n+1}\right]\end{array}$

$\begin{array}{l}\Rightarrow \sum _{k=1}^n t_k=2\left\{\left(\begin{array}{ll}1& \frac{1}{2}\end{array}\right)+\left(\begin{array}{cc}\frac{1}{2}& \frac{1}{3}\end{array}\right)+\cdots +\right.\\ \left.\left(\frac{1}{n}-\frac{1}{n+1}\right)\right\}\\ =2\left(1-\frac{1}{n+1}\right)=\frac{2n}{n+1}\end{array}$