# Evaluating $\sum_{n=0}^{\infty}ne^{1-n}$ using calculus

I'm trying to evaluate the following integral which popped up in MIT Integration Bee 2015 which involves the floor function.

$$\int_{0}^{\infty}\left(xe^{1-x}-\lfloor x\rfloor e^{1-\lfloor x\rfloor}\right)\mathrm dx$$

My Attempt:

\begin{aligned}\mathrm I &=\int_{0}^{\infty}xe^{1-x}\mathrm dx-\sum_{n=0}^{\infty}\int_{n}^{n+1}ne^{1-n}\mathrm dx\\ &= e-\sum_{n=0}^{\infty}ne^{1-n}=e\biggl(1+\sum_{n=0}^{\infty}\left(e^{1-n}\right)'\biggr)\end{aligned}

I'm getting stuck at this step because I'm not able to figure out how to evaluate this sum. I know one way is to use differentiation to get an expression for the sum but I'm not sure how to proceed. A hint in the right direction would be appreciated. Thanks

Note: This is different from How can I evaluate $$\sum_{0}^{\infty}(n+1)x^n$$? . This problem is about bringing the sum into the form from which differentiation would yield the result.

• Note that$$\sum_{n=0}^\infty ne^{1 - n} = \sum_{m=-1}^\infty (m + 1)e^{-m} = \sum_{m=0}^\infty (m + 1)(e^{-1})^m.$$ – Theo Bendit May 15 '19 at 6:06
• You can't differentiate with respect to $n$, Paras. But you can differentiate $\sum_ne^{1-n}x^n$ with respect to $x$. – Gerry Myerson May 15 '19 at 6:23
• Precisely and then setting $x=1$ gives the result. Thanks @GerryMyerson :) – Paras Khosla May 15 '19 at 6:27
• Ah. Why did this not occur to me. Thanks @TheoBendit :) – Paras Khosla May 15 '19 at 6:29
• So, if you can do the problem now, Paras, let me encourage you to write up and post an answer. – Gerry Myerson May 15 '19 at 6:31

Thanks to motivation from @Gerry Myerson. I'm attempting to answer my own question. Let the integral in question be denoted by $$\mathrm I$$ .\begin{aligned}\mathrm I &=\int_{0}^{\infty}xe^{1-x}\mathrm dx-\sum_{n=0}^{\infty}\int_{n}^{n+1}ne^{1-n}\mathrm dx\\ &= e-\sum_{n=0}^{\infty}ne^{1-n}=e-\sum_{n=0}^{\infty}\left(e^{1-n}x^n\right)'\end{aligned}
For the sum, differentiating $$\sum_{n=0}^{\infty}e^{1-n}x^n$$ term-by-term is the way to go. $$e\sum_{n=0}^{\infty}\left(\dfrac{x}{e}\right)^n=\dfrac{e^2}{e-x}\implies \sum_{n=0}^{\infty}ne^{1-n}x^{n-1}=\dfrac{e^2}{(e-x)^2}$$
Plugging in $$x=1$$ and consequently the value of the infinite sum into the value of the expression for the integral yields: $$\mathrm I =e-\dfrac{e^2}{(e-1)^2}=\dfrac{e^3-3e^2+e}{e^2-2e+1}$$