It can be shown that $\int _ { 0 } ^ { \infty } x e ^ { - x } d x$ = 1. Use this fact and the integral test to construct an appropriate convergent infinite series. Is $\sum _ { \mathrm { k } = 0 } ^ { \infty } \mathrm { e } ^ { - \mathrm { k } }$ the correct series?

Question

It can be shown that

\int _ { 0 } ^ { \infty } x e ^ { - x } d x

= 1. Use this fact and the integral test to construct an appropriate convergent infinite series. Is

\sum _ { \mathrm { k } = 0 } ^ { \infty } \mathrm { e } ^ { - \mathrm { k } }

the correct series?

Quizplus · Accepted Answer

The integral $\int _ { 0 } ^ { \infty } x e ^ { - x } d x = 1$ suggests using the integral test on a series that involves $x e^{-x}$. However, the series $\sum _ { k = 0 } ^ { \infty } e ^ { - k }$ corresponds to the integral $\int _ { 0 } ^ { \infty } e ^ { - x } d x$, not $x e ^ { - x } d x$. The correct series that would be analogous to the given integral and converge to a similar form would involve the term $ke^{-k}$, such as $\sum _ { k = 1 } ^ { \infty } k e ^ { - k }$, because it more closely matches the integrand $x e^{-x}$.

It Can Be Shown That ∫0∞xe−xdx\int _ { 0 } ^ { \infty } x e ^ { - x } d x∫0∞​xe−xdx

It Can Be Shown That $\int _ { 0 } ^ { \infty } x e ^ { - x } d x$