The sum of the series

Find the value of sum of

$\displaystyle \frac{1}{{11}}+\frac{1}{{11+22}}+\frac{1}{{11+22+33}}+...+\frac{1}{{11+22+33+...+2013}}$ .

Solution

$\displaystyle \frac{1}{{11}}+\frac{1}{{11+22}}+\frac{1}{{11+22+33}}+...+\frac{1}{{11+22+33+...+2013}}$

$\displaystyle =\frac{1}{{11}}\ \left[ {1+\frac{1}{{1+2}}+\frac{1}{{1+2+3}}+...+\frac{1}{{1+2+3+...+183}}} \right]$ 

$\displaystyle =\frac{1}{{11}}\ \left[ {1+\frac{1}{3}+\frac{1}{6}+...+\frac{1}{{\frac{{183}}{2}(1+183)}}} \right]$ 

$\displaystyle =\frac{1}{{11}}\left[ {1+\frac{1}{3}+\frac{1}{6}+...+\frac{2}{{183\times 184}}} \right]$ 

$\displaystyle =\frac{2}{{11}}\left[ {\frac{1}{2}+\frac{1}{6}+\frac{1}{{12}}+...+\frac{1}{{183\times 184}}} \right]$

$\displaystyle =\frac{2}{{11}}\left[ {\frac{1}{{1\times 2}}+\frac{1}{{2\times 3}}+\frac{1}{{3\times 4}}+...+\frac{1}{{183\times 184}}} \right]$ 

$\displaystyle =\frac{2}{{11}}\ \left[ {\left( {\frac{1}{1}-\frac{1}{2}} \right)+\left( {\frac{1}{2}-\frac{1}{3}} \right)+\left( {\frac{1}{3}-\frac{1}{4}} \right)+...+\left( {\frac{1}{{183}}-\frac{1}{{184}}} \right)} \right]$ 

$\displaystyle =\frac{2}{{11}}\left( {1-\frac{1}{{184}}} \right)$ 

$\displaystyle =\frac{2}{{11}}\times \frac{{183}}{{184}}$ 

$\displaystyle =\frac{{183}}{{1012}}$