Use the integral test to prove the following so-called -series test. The series

$\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{1}}^{\mathbf{\infty }}\frac{\mathbf{1}}{{\mathbf{n}}^{\mathbf{p}}}$isrole="math" localid="1664249612966" $$\begin{gathered} \left\{\mathrm{convergent} \mathrm{if} p>1 , \\ \mathrm{divergent} \mathrm{if} p⩽1.\right\} \end{gathered}$$

If the partial sum${\mathbf{\text{S}}}_{\mathbf{n}}$ of an infinite series tend to a limit S, then the series is called convergent.

If the partial sum${\mathbf{\text{S}}}_{\mathbf{n}}$ of an infinite series do not approach to a limit, the series is called divergent.

The limiting value S is called the sum of the series.

The given series is $\sum _{n=1}^{\infty }\frac{1}{n^p}$.

Use the integral in the given series $\underset{1}{\overset{\infty }{\int }}\frac{1}{n^p} dn$.

Solve the integral for$p>1$ as:

If $p=2$, then:

$\underset{1}{\overset{\infty }{\int }}\frac{1}{n^2} dn=\underset{1}{\overset{\infty }{\int }}{n}^{-2} dn$

Now, solve the integral is as follows:

Use the integral formula $\int x^a dx=\frac{x^{a+1}}{a+1} dx+c$, where is a constant.

$\begin{array}{rcl}\underset{1}{\overset{\infty }{\int }}{n}^{-2} dn& =& {\left[\frac{n^{-2+1}}{-2+1}\right]}_1^{\infty }\\ & =& {\left[\frac{n^{-1}}{-1}\right]}_1^{\infty }\\ & =& -{\left[\frac{1}{n}\right]}_1^{\infty }\\ & =& -\left[\frac{1}{\infty }-\frac{1}{1}\right]\\ & =& -\left[0-1\right]\\ & =& 1\end{array}$

Here, the integral value is a finite value, that is 1.

Therefore, the given series converges for $p=2$.

If $p=3$, then:

$\underset{1}{\overset{\infty }{\int }}\frac{1}{n^3} dn=\underset{1}{\overset{\infty }{\int }}{n}^{-3} dn$

Solve the integral is as follows:

$\begin{array}{rcl}\underset{1}{\overset{\infty }{\int }}{n}^{-3} dn& =& {\left[\frac{n^{-3+1}}{-3+1}\right]}_1^{\infty }\\ & =& {\left[\frac{n^{-2}}{-2}\right]}_1^{\infty }\\ & =& -{\left[\frac{1}{2n^2}\right]}_1^{\infty }\\ & =& -\left[\frac{1}{\infty }-\frac{1}{2}\right]\\ & =& \frac{1}{2}\end{array}$

Here, the integral value is a finite value, that is $\frac{1}{2}$. Therefore, the given series converges for $p=3$.

Now, solve the integral for $p⩽1$:

If $p=0$:

$\underset{1}{\overset{\infty }{\int }}\frac{1}{n^0} dn=\underset{1}{\overset{\infty }{\int }}1 dn$

Solve the integral is as follows:

role="math" localid="1664251356093" $\begin{array}{rcl}\underset{1}{\overset{\infty }{\int }}1 dn& =& {\left[n\right]}_1^{\infty }\\ & =& \infty -1\\ & =& \infty \end{array}$

Here, the integral value is infinite.

Therefore, the given series diverges for $p=0$.

If $p=-1$:

$\underset{1}{\overset{\infty }{\int }}\frac{1}{n^{-1}} dn=\underset{1}{\overset{\infty }{\int }}n dn$

Solve the integral is as follows:

$\begin{array}{rcl}\underset{1}{\overset{\infty }{\int }}n dn& =& {\left[\frac{n^2}{2}\right]}_1^{\infty }\\ & =& \left[\frac{\infty }{2}-\frac{1}{2}\right]\\ & =& \infty \end{array}$

Here, the integral value is an infinite.

Therefore, the given series diverges for $p=-1$.

Hence, it is proved that by using the integral test, the series$\sum _{n=1}^{\infty }\frac{1}{n^p}$ is convergent for$p>1$ and the series is divergent for $p⩽1$.