Question

Use the integral test to show that$\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{0}}^{\mathbf{\infty }}{\mathit{e}}^{\mathbf{-}{\mathbf{n}}^{\mathbf{2}}}$ converges.

Short answer

The series$\sum _{n=0}^{\infty }{e}^{-n^2}$ is convergent.

Step-by-step solution

Definition of Gaussian integral, convergent, and divergent

• The Gaussian integral which is also known as Poisson integral. The area of Gaussian function ${\mathbf{\text{f(x)=e}}}^{\mathbf{-}{\mathbf{x}}^{\mathbf{2}}}$and over the x-axis is as follows:

$\underset{\mathbf{-}\mathbf{\infty }}{\overset{\mathbf{\infty }}{\mathbf{\int }}}{\mathbf{e}}^{\mathbf{-}{\mathbf{x}}^{\mathbf{2}}}\mathbf{dx}\mathbf{=}\sqrt{\mathbf{\pi }}$

Then, half of the area of Gaussian function and -axis is: $\underset{\mathbf{0}}{\overset{\mathbf{\infty }}{\mathbf{\int }}}{\mathit{e}}^{\mathbf{-}{\mathbf{x}}^{\mathbf{2}}}\mathit{d}\mathit{x}\mathbf{=}\frac{\sqrt{\mathbf{\pi }}}{\mathbf{2}}$.

• If the partial sum localid="1664193252645" ${\mathbf{\text{S}}}_{\mathbf{n}}$of an infinite series tend to a limit Sthen the series is called convergent. If the partial sums ${\mathbf{\text{S}}}_{\mathbf{n}}$ of an infinite series do not approach a limit, the series is called divergent. The limiting value S is called the sum of the series.

Apply the integral test

Given that the limit is$\sum _{n=0}^{\infty }{e}^{-n^2}$.

Now, apply the integral test as:

$\underset{0}{\overset{\infty }{\int }}{e}^{-n^2}dn$

This is half of the area of Gaussian function and x-axis which is, $\underset{0}{\overset{\infty }{\int }}{e}^{-n^2}dn=\frac{\sqrt{\pi }}{2}$.

The integral value is finite, therefore the given series$\sum _{n=0}^{\infty }{e}^{-n^2}$converges.

Solve the integral

Now, compare with the integral$\underset{}{\overset{\infty }{\int }}{e}^{-n}dn$.

Evaluate the integral is as follows:

$$\begin{gathered} \underset{}{\overset{\infty }{\int }}{e}^{-n}dn={\left[\frac{e^{-n}}{-1}\right]}^{\infty } \\ =-\left[e^{-\infty }\right] \\ =0 \end{gathered}$$

The integral value is finite, therefore the integral$\underset{}{\overset{\infty }{\int }}{e}^{-n}dn$ converges. Here, both the integral converges.

Hence, it is proved that series$\sum _{n=0}^{\infty }{e}^{-n^2}$ converges.