Answer
$$\frac{{{e^2}}}{{e + 1}}$$
Work Step by Step
$$\eqalign{
& e - 1 + \frac{1}{e} - \frac{1}{{{e^2}}} + \cdots \cr
& {\text{Sum of a Geometric Series }}\left( {{\text{see page 635}}} \right) \cr
& {\text{The infinite geometric series}} \cr
& a + ar + a{r^2} + a{r^3} + \cdots \cr
& {\text{converges}}{\text{, if }}r{\text{ is in }}\left( { - 1,1} \right),{\text{ to the sum }}\frac{a}{{1 - r}}.{\text{ And diverges if }}r{\text{ is not in }}\left( { - 1,1} \right) \cr
& {\text{then this is a geometric series}}{\text{, with }}a = {a_1} = 4/5 \cr
& r = \frac{{ - 1}}{e} = - \frac{1}{e} \cr
& {\text{Since }}r{\text{ is in the interval }}\left( { - 1,1} \right),{\text{ the series converges and has a sum }}\frac{a}{{1 - r}} \cr
& \frac{a}{{1 - r}} = \frac{e}{{1 - \left( { - 1/e} \right)}} \cr
& {\text{simplifying}} \cr
& \frac{a}{{1 - r}} = \frac{e}{{\left( {e + 1} \right)/e}} \cr
& \frac{a}{{1 - r}} = \frac{{{e^2}}}{{e + 1}} \cr} $$
