Question

Evaluate: $\sum _{r=1}^{\infty }\frac{(1+a+a^2+\dots a^{r-1})}{r!}$

Answer 1

Simplifying the given expression:

$\sum _{r=1}^{\infty }\left(\frac{1+a+a^2+\dots a^{r-1}}{r!}\right)$

Here we can see that numerator is a $n^{th}$ term G.P. with common ratio $a$ and first term $1$

Therefore,

$$\begin{gathered} \Rightarrow \sum _{r=1}^{\infty }\left(\frac{{\displaystyle \frac{1(a^r-1)}{(a-1)}}}{r!}\right)\left[\text{Using sum of}{n}^{th}\text{term of}G.P.\right] \\ \Rightarrow \left(\frac{1}{a-1}\right)\left[\sum _{r=1}^{\infty }\left(\frac{a^r}{r!}\right)-\sum _{r=1}^{\infty }\left(\frac{1}{r!}\right)\right] \\ \Rightarrow \left(\frac{1}{a-1}\right)\left[\left(\left(\frac{a}{1!}\right)+\left(\frac{a^2}{2!}\right)+...\right)-\left(\left(\frac{1}{1!}\right)+\left(\frac{1}{2!}\right)+...\right)\right]\left[\text{expanding}\right] \\ \Rightarrow \left(\frac{1}{a-1}\right)\left[e^a-1-(e-1)\right]\left[\sum _{r=1}^{\infty }\left(\frac{a^r}{r!}\right)=e^a-1and\sum _{r=1}^{\infty }\left(\frac{1}{r!}\right)=e-1\right] \\ \Rightarrow \frac{e^a-e}{a-1} \end{gathered}$$

Therefore, $\sum _{r=1}^{\infty }\frac{(1+a+a^2+\dots a^{r-1})}{r!}=\frac{e^a-e}{a-1}$