I am solving a question related to Laplace transform and I get the following sum: $$\sum_{n=1}^{\infty}e^{-sn}\left(\frac{1}{1+s}\right)^{n}.$$ The question asks me to show that this is equal to $\frac{e^{-s}}{1+s-e^{-s}}$. I recognise the possibility to use the geometric sum: \begin{align} \sum_{n=1}^{\infty}e^{-sn}\left(\frac{1}{1+s}\right)^{n}&=\sum_{n=1}^{\infty}\left(\frac{1}{(1+s)e^{s}}\right)^{n}\\&=\frac{1}{1-\dfrac{1}{(1+s)e^{s}}}\\&=\frac{e^{s}(1+s)}{e^{s}(1+s)-1}. \end{align} By multiplying the fraction by $\frac{e^{-s}}{e^{-s}}$ I can get, $$\sum_{n=1}^{\infty}e^{-sn}\left(\frac{1}{1+s}\right)^{n}=\frac{1+s}{1+s-e^{-s}}. $$ Therefore, my numerator is different to the numerator I am asked to show. Can someone explain to me where I went wrong?
Edit:
Based on the comment I solved it. \begin{align} \sum_{n=1}^{\infty}e^{-sn}\left(\frac{1}{1+s}\right)^{n}&=\frac{e^{-s}}{1+s}\sum_{n=0}^{\infty}\left(\frac{1}{(1+s)e^{s}}\right)^{n}\\&=\frac{e^{-s}}{1+s}\cdot\frac{e^{s}(1+s)}{e^{s}(1+s)-1}. \end{align} Which can be simplified and again by multiplying by $\frac{e^{-s}}{e^{-s}}$ I do get, \begin{equation} \sum_{n=1}^{\infty}e^{-sn}\left(\frac{1}{1+s}\right)^{n}=\frac{e^{-s}}{1+s-e^{-s}} \end{equation}
Hint:
It is IGP from $n=1$ to $\infty$. The sum is $\frac{r}{1-r}$, here $r=\frac{e^{-s}}{1+s}$.