Sum of an integral

Calculus Level 3

n = 1 n n + 1 x e x d x = ? \large \sum_{n=1}^\infty \int_n^{n+1}xe^{-x}dx = \ ?

Notation: e 2.718 e \approx 2.718 denotes the Euler's number .

6 e \frac 6e 2 e \frac 2e 8 e \frac 8e 4 e \frac 4e

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1 solution

S = n = 1 n n + 1 x e x d x By integration by parts = n = 1 ( x e x n n + 1 + n n + 1 e x d x ) = n = 1 [ x + 1 e x ] n + 1 n = n = 1 ( n + 1 e n n + 2 e n + 1 ) = 2 e \begin{aligned} S & = \sum_{n=1}^\infty \int_n^{n+1} xe^{-x} dx & \small \color{#3D99F6} \text{By integration by parts} \\ & = \sum_{n=1}^\infty \left(- xe^{-x} \bigg|_n^{n+1} + \int_n^{n+1} e^{-x} dx \right) \\ & = \sum_{n=1}^\infty \left[ \frac {x+1}{e^x} \right]_{n+1}^n \\ & = \sum_{n=1}^\infty \left(\frac {n+1}{e^n} - \frac {n+2}{e^{n+1}} \right) \\ & = \boxed{\dfrac 2e} \end{aligned}

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