# How do you use the Integral Test to determine convergence or divergence of the series: sum n e^-n from n=1 to infinity?

Jul 19, 2015

Take the integral ${\int}_{1}^{\infty} x {e}^{-} x \mathrm{dx}$, which is finite, and note that it bounds ${\sum}_{n = 2}^{\infty} n {e}^{- n}$. Therefore it is convergent, so ${\sum}_{n = 1}^{\infty} n {e}^{- n}$ is as well.

#### Explanation:

The formal statement of the integral test states that if $f \in \left[0 , \infty\right) \rightarrow \mathbb{R}$ a monotone decreasing function which is non-negative. Then the sum ${\sum}_{n = 0}^{\infty} f \left(n\right)$ is convergent if and only if ${\text{sup}}_{N > 0} {\int}_{0}^{N} f \left(x\right) \mathrm{dx}$ is finite. (Tau, Terence. Analysis I, second edition. Hindustan book agency. 2009).

This statement may seem a bit technical, but the idea is the following. Taking in this case the function $f \left(x\right) = x {e}^{- x}$, we note that for $x > 1$, this function is decreasing. We can see this by taking the derivative. $f ' \left(x\right) = {e}^{- x} - x {e}^{- x} = \left(1 - x\right) {e}^{- x} < 0$, since $x > 1$, so $\left(1 - x\right) < 0$ and ${e}^{- x} > 0$.

Due to this, we note that for any $n \in {\mathbb{N}}_{\ge 2}$ and $x \in \left[1 , \infty\right)$ such that $x \le n$ we have $f \left(x\right) \ge f \left(n\right)$. Therefore ${\int}_{n - 1}^{n} f \left(x\right) \mathrm{dx} \ge {\int}_{n - 1}^{n} f \left(n\right) \mathrm{dx} = f \left(n\right)$, so ${\sum}_{n = 1}^{N} f \left(n\right) \le f \left(1\right) + {\sum}_{n = 2}^{N} {\int}_{n - 1}^{n} f \left(x\right) \mathrm{dx} = f \left(1\right) + {\int}_{1}^{N} f \left(x\right) \mathrm{dx}$.

${\int}_{1}^{\infty} f \left(x\right) \mathrm{dx} = {\int}_{1}^{\infty} x {e}^{- x} \mathrm{dx} = - {\int}_{x = 1}^{\infty} x {\mathrm{de}}^{- x} = - x {e}^{- x} {|}_{1}^{\infty} + {\int}_{1}^{\infty} {e}^{- x} \mathrm{dx}$$= - x {e}^{- x} - {e}^{- x} {|}^{\infty} _ 1 = \frac{2}{e}$ using integration by parts and that ${\lim}_{x \rightarrow \infty} {e}^{-} x = {\lim}_{x \rightarrow \infty} x {e}^{-} x = 0$.

Since $f \left(x\right) \ge 0$, we have $\frac{e}{2} = {\int}_{1}^{\infty} f \left(x\right) \mathrm{dx} \ge {\int}_{1}^{N} f \left(x\right) \mathrm{dx}$, so ${\sum}_{n = 1}^{N} f \left(n\right) \le f \left(1\right) + \frac{2}{e} = \frac{3}{e}$. Since $f \left(n\right) \ge 0$, the series ${\sum}_{n = 1}^{N} f \left(n\right)$ increases as $N$ increases. Since it is bounded by $\frac{3}{e}$, it must converge. Therefore ${\sum}_{n = 1}^{\infty} f \left(n\right)$ converges.