Question

How do you use the integral test to determine if $\sum{{{n}^{k}}{{e}^{-n}}}$ from $\left[ 1,\infty \right)$ where k is an integer is convergent or divergent?

Answer 1

From the given question, we can see that the function is always positive since the given interval is $\left[ 1,\infty \right).$ We can solve this question by first replacing n by x in the test function and then differentiating the function with respect to x to know if the function is increasing or decreasing. We then integrate the function considering the critical points. If the integral value is positive, we can say that the function is convergent, else it will be divergent.

Complete step-by-step answer:
We need to use the integral test to determine if the given function is convergent or divergent. The function is $f\left( n \right)=\sum{{{n}^{k}}{{e}^{-n}}}.$ We first differentiate the function. In order to do so, we need to replace n by x in the function.
Let $f\left( x \right)=\sum{{{x}^{k}}{{e}^{-x}}}.$
Now let us differentiate this function,
$\Rightarrow \dfrac{d}{dx}\left( f\left( x \right) \right)=\dfrac{d}{dx}\left( {{x}^{k}}{{e}^{-x}} \right)$
To differentiate this, we use the multiplication rule of differentiation, which is given as follows,
$\dfrac{d}{dx}\left( uv \right)=u\dfrac{dv}{dx}+v\dfrac{du}{dx}\ldots \ldots \left( 1 \right)$
We apply this rule as shown.
$\Rightarrow \dfrac{d}{dx}\left( {{x}^{k}}{{e}^{-x}} \right)={{x}^{k}}\dfrac{d}{dx}\left( {{e}^{-x}} \right)+{{e}^{-x}}\dfrac{d}{dx}\left( {{x}^{k}} \right)\ldots \ldots \left( 2 \right)$
Simplifying the first term and since $\dfrac{d}{dx}\left( {{e}^{-x}} \right)=-{{e}^{-x}},$
$\Rightarrow {{x}^{k}}\dfrac{d}{dx}\left( {{e}^{-x}} \right)={{x}^{k}}.\left( -{{e}^{-x}} \right)$
Simplifying the next term and since $\dfrac{d}{dx}\left( {{x}^{k}} \right)=k.\left( {{x}^{k-1}} \right),$
$\Rightarrow {{e}^{-x}}\dfrac{d}{dx}\left( {{x}^{k}} \right)={{e}^{-x}}.\left( k.{{x}^{k-1}} \right)$
Putting together both the simplified terms in the equation (2),
$\Rightarrow \dfrac{d}{dx}\left( {{x}^{k}}{{e}^{-x}} \right)=-{{x}^{k}}.{{e}^{-x}}+k.{{e}^{-x}}.{{x}^{k-1}}$
Taking the common terms out, that is ${{x}^{k-1}}$ and ${{e}^{-x}},$
$\Rightarrow \dfrac{d}{dx}\left( {{x}^{k}}{{e}^{-x}} \right)={{x}^{k-1}}{{e}^{-x}}\left( -x+k \right)$
As per the given interval of $\left[ 1,\infty \right),$ by substituting the value of x from here, we see that the function is decreasing.
We now have to integrate the function $f\left( x \right)$ within the limits $\left[ 1,\infty \right).$
Let $I=\int\limits_{1}^{\infty }{{{x}^{k}}{{e}^{-x}}dx}.$
Now we need to find a general solution for this integral. Let us assume k=1 for the first case as shown.
$\Rightarrow {{I}_{1}}=\int\limits_{1}^{\infty }{x{{e}^{-x}}dx}$
We all know integration by parts method which is given as,
$\int{udv=uv-\int{vdu\ldots \ldots \left( 3 \right)}}$
Where $u=x$ and $dv={{e}^{-x}}dx.$
Differentiating both sides of first equation we get $du=dx\ldots \ldots \left( 4 \right)$
To find v now, we integrate both sides of the second equation which gives,
$\Rightarrow \int{dv=\int{{{e}^{-x}}dx}}$
Since we know that $\int{{{e}^{-x}}dx}=-{{e}^{-x}},$ we get
$\Rightarrow v=-{{e}^{-x}}\ldots \ldots \left( 5 \right)$
Now we substitute these values in equation (3),
$\Rightarrow {{I}_{1}}=\left[ -x{{e}^{-x}} \right]_{1}^{\infty }+\int\limits_{1}^{\infty }{{{e}^{-x}}dx}$
Substituting the upper and lower limits, we get
$\Rightarrow {{I}_{1}}=-\left[ \infty .{{e}^{-\infty }}-1{{e}^{-1}} \right]+\left[ -{{e}^{-x}} \right]$
We know that ${{e}^{-\infty }}=0,$
$\Rightarrow {{I}_{1}}=\left[ \dfrac{1}{e} \right]-\left[ {{e}^{-\infty }}-{{e}^{-1}} \right]$
Simplifying this by adding the remaining terms,
$\begin{aligned} & \Rightarrow {{I}_{1}}=\left[ \dfrac{1}{e} \right]+\left[ \dfrac{1}{e} \right] \\\ & \Rightarrow {{I}_{1}}=\left[ \dfrac{2}{e} \right] \\\ \end{aligned}$
Next let us assume k=2,
$\Rightarrow {{I}_{2}}=\int\limits_{1}^{\infty }{{{x}^{2}}{{e}^{-x}}dx}$
Here, $u={{x}^{2}}$ and $dv={{e}^{-x}}dx.$
Also, $du=2xdx$ and $v=-{{e}^{-x}}.$
Integrating this by parts similar to the previous methods,
$\Rightarrow {{I}_{2}}=\left[ -x{{e}^{-x}} \right]_{1}^{\infty }+2\int\limits_{1}^{\infty }{x{{e}^{-x}}dx}$
Substituting the limits and the second term as ${{I}_{1}},$
$\Rightarrow {{I}_{2}}=-\left[ \infty .{{e}^{-\infty }}-1{{e}^{-1}} \right]+2{{I}_{1}}$
Simplifying the result by evaluating further,
$\Rightarrow {{I}_{2}}=\dfrac{1}{e}+2{{I}_{1}}$
We know that ${{I}_{1}}=\dfrac{2}{e},$ therefore,
$\begin{aligned} & \Rightarrow {{I}_{2}}=\dfrac{1}{e}+\dfrac{4}{e} \\\ & \Rightarrow {{I}_{2}}=\dfrac{5}{e} \\\ \end{aligned}$
So, in a general form,
$\Rightarrow {{I}_{k}}=\int\limits_{1}^{\infty }{{{x}^{k}}{{e}^{-x}}dx}$
Expanding this using integration by parts, we get
$\Rightarrow {{I}_{k}}=\left[ -x{{e}^{-x}} \right]_{1}^{\infty }+k\int\limits_{1}^{\infty }{{{x}^{k-1}}{{e}^{-x}}dx}$
This can be simplified further as,
$\Rightarrow {{I}_{k}}=\dfrac{1}{e}+k{{I}_{k-1}}$
Now we can expand this as follows,
$\Rightarrow {{I}_{k}}=\dfrac{1}{e}+k\left( \dfrac{1}{e}+\left( k-1 \right) \right){{I}_{k-2}}$
We continue to expand like this and taking the sum of all the terms, we get
$\Rightarrow {{I}_{k}}=\dfrac{1+k+k\left( k-1 \right)+\ldots +k!}{e}$
This can be written in summation form as,
$\Rightarrow {{I}_{k}}=\dfrac{1}{e}\sum\limits_{i=1}^{k}{\dfrac{k!}{i!}}$
This equation shows that,
$\int\limits_{1}^{\infty }{{{x}^{k}}{{e}^{-x}}dx}=\dfrac{1}{e}\sum\limits_{i=1}^{k}{\dfrac{k!}{i!}}$
This means that the above equation is convergent for any k and therefore, $\sum{{{n}^{k}}{{e}^{-n}}}$ is also convergent.
Hence, we have shown that $\sum{{{n}^{k}}{{e}^{-n}}}$ is convergent for k being any integer in the interval $\left[ 1,\infty \right).$

Note: While solving this question, the students need to be careful while performing integration and differentiation as there must be a thorough understanding in these concepts in order to solve this question easily. Also, care must be taken while substituting the limits during integration and we need to first substitute the upper limit followed by the lower limit.