Can the series $\sum_{r=0}^{\infty}\frac{e^{-r}}{1+r}$ be further simplified?

Question

I came across this series at work the other day and was wondering if it can be simplified in an explicit closed form without summations? I tried expanding the denominator as an infinite series and viewing it in terms of arithmetico-geometric series, but it got me nowhere. The series is:

$$\sum_{r=0}^{\infty}\frac{e^{-r}}{1+r}$$

My instinct tells me you can't further simplify this but I'd appreciate some help verifying or denying this.

Answer 1

Start with the series $$\dfrac{1}{1-t} = \displaystyle\sum_{n = 0}^{\infty}t^n,$$ and integrate term-by-term from $t = 0$ to $t = x$ to get $$-\ln(1-x) = \displaystyle\sum_{n = 0}^{\infty}\dfrac{x^{n+1}}{n+1}.$$ Then, divide by $x$ to get $$-\dfrac{\ln(1-x)}{x} = \displaystyle\sum_{n = 0}^{\infty}\dfrac{x^{n}}{n+1}.$$ Finally, plug in $x = e^{-1}$ to get $$\displaystyle\sum_{n = 0}^{\infty}\dfrac{e^{-n}}{n+1} = -e\ln(1-e^{-1}).$$

Answer 2

Let $f(x)$ be given by the series

$$f(x)=\sum_{n=0}^\infty \frac{x^{n+1}}{1+n}\tag 1$$

for $|x|<1$.

Now differentiating $(1)$ reveals

$$f'(x)=\frac{1}{1-x}\tag2$$

whereupon integrating $(2)$ shows

$$f(e^{-1})-f(0)=\int_0^{e^{-1}}\frac{1}{1-x}\,dx$$

Can you wrap this up now?