Question

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

Answer 1

Hint: From the given function, we find that the function is always positive since the interval is \[\left[ 1,\infty \right)\]. We have to differentiate the given function with respect to x after replacing n by x to know whether the function is decreasing or not. We find that there is one critical point in the function that is 1. Then integrate the function considering the critical point. If the integration is more than 0, then the function is convergent.

Complete step by step answer:
According to the question, we are asked to find whether the function is convergent or divergent.
We have been given the function is \[f\left( n \right)=n{{e}^{-n}}\] from n=1 to infinity that is, in the interval \[\left[ 1,\infty \right)\].--------(1)
First, we have to find whether the function is decreasing or increasing.
So we have to differentiate the function.
Let us now replace n with x in the function (1) to differentiate with respect to x.
That is, we have to differentiate \[x{{e}^{-x}}\] with respect to x.
Let \[f\left( x \right)=x{{e}^{-x}}\].
\[\dfrac{d}{dx}\left( f\left( x \right) \right)=\dfrac{d}{dx}\left( x{{e}^{-x}} \right)\]
We have to use multiplication rule of differentiation, that is, \[\dfrac{d}{dx}\left( uv \right)=u\dfrac{dv}{dx}+v\dfrac{du}{dx}\].
Here, u=x and \[v={{e}^{-x}}\].
We get,
\[\dfrac{d}{dx}\left( x{{e}^{-x}} \right)=x\dfrac{d}{dx}\left( {{e}^{-x}} \right)+{{e}^{-x}}\dfrac{dx}{dx}\]
We know that, \[\dfrac{dx}{dx}=1\]. We get
\[\dfrac{d}{dx}\left( x{{e}^{-x}} \right)=x\dfrac{d}{dx}\left( {{e}^{-x}} \right)+{{e}^{-x}}\]
Since \[\dfrac{d}{dx}\left( {{e}^{-x}} \right)=-{{e}^{-x}}\], we get
\[\dfrac{d}{dx}\left( x{{e}^{-x}} \right)=x\left( -{{e}^{-x}} \right)+{{e}^{-x}}\]
On taking the common term, that is, \[{{e}^{-x}}\], we get
\[\dfrac{d}{dx}\left( x{{e}^{-x}} \right)={{e}^{-x}}\left( 1-x \right)\].
Here, the interval is \[\left[ 1,\infty \right)\].
On substituting the value of x from the interval, we find that the value decreases consecutively.
Therefore, the function is decreasing.
Now, let us run the Integral Test.
That is, we have to integrate the function f(x) with respect to x under the limits \[\left[ 1,\infty \right)\].
Therefore,
\[\int\limits_{1}^{x}{x{{e}^{-x}}dx=\displaystyle \lim_{t\to \infty }}\int\limits_{1}^{t}{x{{e}^{-x}}dx}\]
We have to use integration by parts, that is, \[\int{udv=uv-vdu}\].
Here, u=x and \[dv={{e}^{-x}}dx\].
Therefore, du=dx.
Now we have to find v.
Integrate \[dv={{e}^{-x}}dx\] on both the sides to get the value of v.
\[\int{dv}=\int{{{e}^{-x}}dx}\]
We know that \[\int{{{e}^{-x}}dx}=-{{e}^{-x}}\]. We get
\[\int{dv}=-{{e}^{-x}}\]
\[\therefore v=-{{e}^{-x}}\]
Now, substitute these values in integration by parts method. We get,
\[\int\limits_{1}^{x}{x{{e}^{-x}}dx=}\displaystyle \lim_{x \to \infty }\left[ \left[ x\left( -{{e}^{-x}} \right) \right]_{1}^{x}-\int\limits_{1}^{x}{-{{e}^{-x}}dx} \right]\]
We know that \[\int{{{e}^{-x}}dx}=-{{e}^{-x}}\]. We get
\[\int\limits_{1}^{x}{x{{e}^{-x}}dx=}\displaystyle \lim_{x \to \infty }\left[ \left[ x\left( -{{e}^{-x}} \right) \right]_{1}^{x}-\left[ -{{e}^{-x}} \right]_{1}^{x} \right]\]
Now, let us substitute the limits in the integral.
\[\int\limits_{1}^{x}{x{{e}^{-x}}dx=}\displaystyle \lim_{x \to \infty }\left[ \left[ -x{{e}^{-x}}-\left( -{{e}^{-1}} \right) \right]-\left[ -{{e}^{-x}}+{{e}^{-1}} \right] \right]\]
On further simplifications, we get
\[\int\limits_{1}^{x}{x{{e}^{-x}}dx=}\displaystyle \lim_{x \to \infty }\left[ \left[ -x{{e}^{-x}}+{{e}^{-1}} \right]+\left[ {{e}^{-x}}-{{e}^{-1}} \right] \right]\]
\[\Rightarrow \int\limits_{1}^{x}{x{{e}^{-x}}dx=}\displaystyle \lim_{x \to \infty }\left[ -x{{e}^{-x}}+{{e}^{-1}}+{{e}^{-x}}-{{e}^{-1}} \right]\]
Cancelling out the same terms with opposite sign, we get
\[\Rightarrow \int\limits_{1}^{x}{x{{e}^{-x}}dx=}\displaystyle \lim_{x \to \infty }\left[ -t{{e}^{-x}}+{{e}^{-x}} \right]\]
Now, let us substitute the value of x as per the limits.
We get
\[\Rightarrow \int\limits_{1}^{x}{x{{e}^{-x}}dx=}\left[ -\infty {{e}^{-\infty }}+{{e}^{-\infty }} \right]\]
We know that \[{{e}^{-\infty }}=0\].
Therefore, \[\int\limits_{1}^{x}{x{{e}^{-x}}dx=}0\]
Here, the integral test shows zero and on increasing the value of x, we find that the value of function also increases.
In the series\[\sum n{{e}^{-n}}\] , when the value of n increases in the given interval the value of the function also increases.
Since the integral is convergent then, the series must also be convergent by Integral Test.
Therefore, the series \[\sum n{{e}^{-n}}\] from n=1 to infinity is convergent.

Note:
 We should always differentiate the function to find whether the function is decreasing or increasing. Then we have to run the integral test. If the integral is less than infinity, then the given function is divergent. Also, the series with the function under the given limit becomes convergent.