$11+\frac{11}{2}+\frac{11}{3}+\frac{11}{4}+\frac{11}{5}+\frac{11}{6}+\frac{11}{7}+\frac{11}{8}+\frac{11}{9}+\frac{11}{10}+$cdots$={\displaystyle \underset{n=}{\overset{A}{}}}{\displaystyle \underset{n=}{\overset{A}{}}}$

The form $of$ your answer will depend $on$ your choice $of$ the lower $limitof$ summation.