Sum of infinite series

Lets evaluates the infinot sum in the picture: 
\[\sum_{i=1}^{\infty}\frac{x}{(1+i)^k}=x\sum_{i=1}^{\infty}\frac{1}{(1+i)^k}=x (\frac{\frac{1}{1+i}}{1-\frac{1}{1+i}})\]
\[=x (\frac{\frac{1}{1+i}}{\frac{i}{1+i}} )=\frac{x}{i}.\]
I am guessing that $x$ might be a typo and you meant $D$ by $x$. If so, then the infinite sum above will be equal to $D/i$ as you expected. 

Here we have used the formula for the sum of infinite series
\[\sum_{k=1}^{\infty}a^k=\frac{a}{1-a}    \text{for}  \ \ |a|<1.\]