# How do you determine convergence or divergence for the summation of n*e^(-n/2) using the integral test how do you answer?

Jun 30, 2015

The series of sequence ${u}_{n}$ is convergent.

#### Explanation:

We have the series of sequence ${u}_{n} = n {e}^{- \frac{n}{2}} = \frac{n}{\sqrt{{e}^{n}}}$.

In order to use the integral test, $f \left(n\right) = {u}_{n}$ must be positive and decreasing on [p;+oo[.

Let's study the sign of $f \left(n\right)$ :

(Note : ${e}^{- \frac{n}{2}}$ is always positive and it has the value of zero only if $n \to + \infty$; therefore, the sign of $f \left(n\right)$ depends only on the numerator : $n$.)

Thus, $f \left(n\right)$ is positive on ]0; +oo[.

Let's study the slope (the growth) of $f \left(n\right)$, which comes down to study the sign of $f ' \left(n\right)$.

The derivative of a function $f \left(x\right) = g \left(x\right) \cdot h \left(x\right)$ is :

$f ' \left(x\right) = g ' \left(x\right) h \left(x\right) + g \left(x\right) h ' \left(x\right)$.

Therefore :

$f \left(n\right) = n \cdot {e}^{- \frac{n}{2}} = g \left(n\right) \cdot h \left(n\right)$

$f ' \left(n\right) = \left(n\right) ' \cdot {e}^{- \frac{n}{2}} + n \cdot \left({e}^{- \frac{n}{2}}\right) '$

$f ' \left(n\right) = {e}^{- \frac{n}{2}} + n \cdot \left(- \frac{n}{2}\right) ' \cdot \left({e}^{- \frac{n}{2}}\right)$

$f ' \left(n\right) = {e}^{- \frac{n}{2}} + n \cdot \left(- \frac{1}{2}\right) \cdot \left({e}^{- \frac{n}{2}}\right)$

$f ' \left(n\right) = {e}^{- \frac{n}{2}} \cdot \left(1 - \frac{n}{2}\right)$

(Again, the sign of $f ' \left(n\right)$ depends only on the numerator : $1 - \frac{n}{2}$).

Thus, $f \left(n\right)$ is positive and decreasing on [+2;+oo[.

So we can use the integral test.

The series of sequence ${u}_{n}$ converges if ${\int}_{p}^{+ \infty} f \left(x\right) \mathrm{dx}$ exists.

Of course, here $f \left(x\right) = {e}^{- \frac{x}{2}} x$ and $p = 2$.

If we want to calculate the integral, we must firstly find the antiderivative of $f \left(x\right)$.

The antiderivative of a function $f \left(x\right) = g ' \left(x\right) h \left(x\right)$ is :

$F \left(x\right) = g \left(x\right) h \left(x\right) - \int g \left(x\right) h ' \left(x\right) \mathrm{dx}$

We have $f \left(x\right) = {e}^{- \frac{x}{2}} \cdot x = g ' \left(x\right) \cdot h \left(x\right)$

We know that the derivative of ${e}^{- \frac{x}{2}}$ is $- \frac{1}{2} {e}^{- \frac{x}{2}}$.

So we can easily find the antiderivative of $g ' \left(x\right) = {e}^{- \frac{x}{2}}$ :

$g \left(x\right) = - 2 \cdot {e}^{- \frac{x}{2}}$.

So $F \left(x\right) = \left(- 2 \cdot {e}^{- \frac{x}{2}} \cdot x\right) - \left(\int - 2 \cdot {e}^{- \frac{x}{2}} \cdot 1\right)$

Again, we can easily find the antiderivative of $- 2 \cdot {e}^{- \frac{x}{2}}$, which is $4 \cdot {e}^{- \frac{x}{2}}$.

$F \left(x\right) = \left(- 2 \cdot {e}^{- \frac{x}{2}} \cdot x\right) - \left(4 \cdot {e}^{- \frac{x}{2}}\right) = - 2 {e}^{- \frac{x}{2}} \cdot \left(x + 2\right) = - \frac{2 \left(x + 2\right)}{\sqrt{{e}^{x}}}$.

Therefore :

${\int}_{p}^{+ \infty} f \left(x\right) \mathrm{dx} = {\left[F \left(x\right)\right]}_{2}^{+ \infty} = ' ' F \left(+ \infty\right) ' ' - F \left(2\right) = 0 + \frac{2 \left(2 + 2\right)}{\sqrt{{e}^{2}}} = \frac{8}{e}$.

(Note : $' ' F \left(+ \infty\right) ' ' = 0$ since $\sqrt{{e}^{x}}$ increases faster to $+ \infty$ than $x$).

Therefore, the series of sequence ${u}_{n}$ is convergent.