91Ó°ÊÓ

The convergent function is given as:

$\underset{\mathrm{n}=2}{\overset{∞}{∑}}\frac{1}{{\mathrm{n}}^{\frac{3}{2}}}$

The series is given as:

$\frac{1}{\sqrt{8}}+\frac{1}{\sqrt{27}}+\frac{1}{\sqrt{64}}+\frac{1}{\sqrt{125}}⋯>\frac{1}{\sqrt{9}}+\frac{1}{\sqrt{36}}+\frac{1}{\sqrt{81}}+⋯$

which is:

$\frac{1}{3}+\frac{1}{6}+\frac{1}{9}+⋯=\frac{1}{3}(1+\frac{1}{2}+\frac{1}{3}+⋯)$

When the harmonic series diverges, the original series diverges.

Comparing:

$3n=n\sqrt{n}$

The p-series test is given as:

$p>1.\underset{\mathrm{n}=2}{\overset{∞}{∑}}\frac{1}{{\mathrm{n}}^{\mathrm{p}}}$

This is convergent and divergent if$p≤1$.

By comparing$\underset{\mathrm{n}=2}{\overset{∞}{∑}}\frac{1}{{\mathrm{n}}^{\frac{3}{2}}}$withrole="math" localid="1658917993283" $\underset{\mathrm{n}=2}{\overset{∞}{∑}}\frac{1}{{\mathrm{n}}^{\mathrm{p}}}$.

If $p=\frac{3}{2}>1$

Then by the p - series test, the series role="math" localid="1658918474467" $\underset{\mathrm{n}=2}{\overset{8}{∑}}\frac{1}{{\mathrm{n}}^{\frac{3}{2}}}$ will be convergent.

The given series is:

$$\begin{gathered} \frac{1}{\sqrt{8}}+\frac{1}{\sqrt{27}}+\frac{1}{\sqrt{64}}+\frac{1}{\sqrt{125}}..⋯>\frac{1}{\sqrt{9}}+\frac{1}{\sqrt{36}}+\frac{1}{\sqrt{81}}+\frac{1}{\sqrt{144}}+..⋯ \\ =\frac{1}{3}+\frac{1}{6}+\frac{1}{9}+\frac{1}{12}+...⋯ \\ =\frac{1}{3}(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...⋯) \end{gathered}$$

As the harmonic series diverges, the original series will also diverge

Now, the expression is given as:

Suppose,

$3n>n\sqrt{2}$

Then,

role="math" localid="1658919405947" $$\begin{gathered} \frac{1}{3\mathrm{n}}<\frac{1}{\mathrm{n}\sqrt{\mathrm{n}}} \\ =\frac{1}{3}+\frac{1}{6}+\frac{1}{9}+\frac{1}{12}+…….<\frac{1}{2\sqrt{2}}+\frac{1}{3\sqrt{3}}+\frac{1}{4\sqrt{4}}+\frac{1}{5\sqrt{5}}+…….......\left(1\right) \\ =\frac{1}{2\sqrt{2}}+\frac{1}{3\sqrt{3}}+\frac{1}{4\sqrt{4}}+\frac{1}{5\sqrt{5}}+……..>\frac{1}{3}+\frac{1}{6}+\frac{1}{9}+\frac{1}{12}+……. \end{gathered}$$

Now, the series is given as:

role="math" localid="1658919005396" $\frac{1}{3}+\frac{1}{6}+\frac{1}{9}+\frac{1}{12}+⋯…….=\frac{1}{3}∑\frac{1}{n}$

Using p- series test, the given series $∑\frac{1}{n}$ is divergent

Then, using the equation (1), the series is written as:

role="math" localid="1658918985849" $\frac{1}{2\sqrt{2}}+\frac{1}{3\sqrt{3}}+\frac{1}{4\sqrt{4}}+\frac{1}{5\sqrt{5}}+⋯……..=\left(\frac{1}{\mathrm{n}\sqrt{\mathrm{n}}}\right)$

The above series is also divergent by the use of the comparison test.

In the first part of this problem this contradicts that$∑\frac{1}{{\mathrm{n}}^{{\displaystyle \frac{3}{2}}}}$ is convergent.

Therefore, the assuming $3n>n\sqrt{n}$ is false.

Hence, the series is $\underset{\mathrm{n}=2}{\overset{8}{∑}}\frac{1}{{\mathrm{n}}^{\frac{3}{2}}}$ will be convergent.

By substituting $n=16$in the equation$3n>n\sqrt{n}$ , it is find that the equation is not correct as shown below:

$$\begin{gathered} 3×16>16×\sqrt{16} \\ 48>16×4 \\ 48>64 \end{gathered}$$

This is not possible.