# How do you use the integral test to determine if Sigma n^ke^-n from [1,oo) where k is an integer is convergent or divergent?

Feb 9, 2017

The series

${\sum}_{= 1}^{\infty} {n}^{k} {e}^{- n}$

is convergent for any integer $k$

#### Explanation:

Choose as test function:

$f \left(x\right) = {x}^{k} {e}^{- x}$

This function is:

(i) non negative, as $f \left(x\right) > 0$ for $x > 0$

(ii) strictly decreasing in $\left(k , + \infty\right)$ as:

$f ' \left(x\right) = k {x}^{k - 1} {e}^{- x} - {x}^{k} {e}^{- x} = {x}^{k - 1} {e}^{- x} \left(k - x\right) < 0$ for $x > k$

(iii) infinitesimal, as:

${\lim}_{x \to \infty} {x}^{k} {e}^{- x} = 0$

(iv) $f \left(n\right) = {n}^{k} {e}^{- n}$

so the convergence of the series is equivalent to the convergence of the integral:

${\int}_{1}^{\infty} {x}^{k} {e}^{- x} \mathrm{dx}$

We have to find a general formula for this integral. Start from $k = 1$, integrating by parts:

${I}_{1} = {\int}_{1}^{\infty} x {e}^{- x} \mathrm{dx} = - {\int}_{1}^{\infty} x d \left({e}^{- x}\right) = {\left[- x {e}^{- x}\right]}_{1}^{\infty} + {\int}_{1}^{\infty} {e}^{- x} \mathrm{dx} = \frac{1}{e} - {\left[{e}^{- x}\right]}_{1}^{\infty} = \frac{2}{e}$

For $k = 2$

${I}_{2} = {\int}_{1}^{\infty} {x}^{2} {e}^{- x} \mathrm{dx} = - {\int}_{1}^{\infty} {x}^{2} d \left({e}^{- x}\right) = {\left[- {x}^{2} {e}^{- x}\right]}_{1}^{\infty} + 2 {\int}_{1}^{\infty} x {e}^{- x} \mathrm{dx} = \frac{1}{e} + 2 {I}_{1} = \frac{5}{e}$

and in general:

${I}_{k} = {\int}_{1}^{\infty} {x}^{k} {e}^{- x} \mathrm{dx} = - {\int}_{1}^{\infty} {x}^{k} d \left({e}^{- x}\right) = {\left[- {x}^{k} {e}^{- x}\right]}_{1}^{\infty} + k {\int}_{1}^{\infty} {x}^{k - 1} {e}^{- x} \mathrm{dx} = \frac{1}{e} + k {I}_{k - 1}$

I_k= 1/e + kI_(k-1) = 1/e +k(1/e+(k-1))I_(k-2) = ... = (1+k+k(k-1)+...+k!)/e = 1/esum_(j=1)^k (k!)/(j!)

So we can assert that:

int_1^oo x^ke^(-x)dx = 1/esum_(j=1)^k (k!)/(j!)

is convergent for every $k$ and this proves that also:

${\sum}_{= 1}^{\infty} {n}^{k} {e}^{- n}$

is convergent.

Mar 28, 2017

Application of L'Hopital's Rule to the Problem

#### Explanation:

Here is an alternative explanation:

The convergence of the "tail" (the infinitely-many terms at the end) of the series determines whether the series itself converges. We will show that the tail of the series is finite. Therefore the series converges.

Assume first that k = 0.
Then
$\Sigma \frac{{n}^{k}}{{e}^{n}} = \Sigma \left(\frac{1}{{e}^{n}}\right)$, which is a convergent geometric series.

Assume k is a POSITIVE integer.
For all x > 0, ${x}^{k} \mathmr{and} {2}^{x}$ are strictly positive.
By applying L'Hopital's rule k times, we see that ...

lim_(x rarr oo)(x^k/2^x) = (k/ln2)lim_(x rarr oo)(x^(k-1)/2^x)=...=(k!)/(ln2)^klim_(x rarr oo)(1/2^x) = 0 .

Therefore, for x sufficiently large, ${x}^{k} < {2}^{x}$.

For such values of x,

${x}^{k} / \left({e}^{x}\right) < {2}^{x} / \left({e}^{x}\right) = {\left(\frac{2}{e}\right)}^{x}$.

Therefore there exists a whole number, N, so that when n > N, we have
${n}^{k} / \left({e}^{n}\right) < {\left(\frac{2}{e}\right)}^{n}$

Now $\Sigma \left({\left(\frac{2}{e}\right)}^{n}\right)$ is a geometric series that converges since 2 < e.

Therefore, the tail of our series $\Sigma \frac{{n}^{k}}{{e}^{n}}$ is bounded above by the tail of the convergent geometric series. [We can compute the upper limit of that value by using the Geometric Series Theorem: $\Sigma \left({\left(\frac{2}{e}\right)}^{n}\right) = \frac{1}{1 - \frac{2}{e}}$ .]

Consequently, $\Sigma \frac{{n}^{k}}{{e}^{n}}$ converges.

Finally, assume k is a NEGATIVE integer.

If k < 0, then for n greater than or equal to 1, ${n}^{k} < 1$. Therefore, the series as a whole is bounded above by $\Sigma \left(\frac{1}{{e}^{n}}\right)$ and converges by the Comparison Test.