Show that $\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{2}}^{\mathbf{\infty }}\mathbf{}\frac{\mathbf{1}}{{\mathbf{n}}^{\frac{\mathbf{3}}{\mathbf{2}}}}$ is convergent. What is wrong with the following "proof" that it diverges? $\frac{\mathbf{1}}{\sqrt{\mathbf{8}}}\mathbf{+}\frac{\mathbf{1}}{\sqrt{\mathbf{27}}}\mathbf{+}\frac{\mathbf{1}}{\sqrt{\mathbf{64}}}\mathbf{+}\frac{\mathbf{1}}{\sqrt{\mathbf{125}}}\mathbf{+}\mathbf{\cdots }\mathbf{>}\frac{\mathbf{1}}{\sqrt{\mathbf{9}}}\mathbf{+}\frac{\mathbf{1}}{\sqrt{\mathbf{36}}}\mathbf{+}\frac{\mathbf{1}}{\sqrt{\mathbf{81}}}\mathbf{+}\frac{\mathbf{1}}{\sqrt{\mathbf{144}}}\mathbf{+}\mathbf{\cdots }$which is $\frac{\mathbf{1}}{\mathbf{3}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{6}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{9}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{12}}\mathbf{+}\mathbf{\cdots }\mathbf{=}\frac{\mathbf{1}}{\mathbf{3}}\mathbf{}\mathbf{(}\mathbf{1}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{3}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{4}}\mathbf{+}\mathbf{\cdots }\mathbf{)}$ . Since the harmonic series diverges, the original series diverges. Hint: Compare 3 n and$\mathbf{n}\sqrt{\mathbf{n}}\mathbf{.}$

The convergent function is given as:

$\sum _{\mathrm{n}=2}^{\infty }\frac{1}{{\mathrm{n}}^{\frac{3}{2}}}$

The series is given as:

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

which is:

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

When the harmonic series diverges, the original series diverges.

Comparing:

$3n=n\sqrt{n}$

The p-series test is given as:

$p>1.\sum _{\mathrm{n}=2}^{\infty }\frac{1}{{\mathrm{n}}^{\mathrm{p}}}$

This is convergent and divergent if$p\le 1$.

By comparing$\sum _{\mathrm{n}=2}^{\infty }\frac{1}{{\mathrm{n}}^{\frac{3}{2}}}$withrole="math" localid="1658917993283" $\sum _{\mathrm{n}=2}^{\infty }\frac{1}{{\mathrm{n}}^{\mathrm{p}}}$.

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

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

The given series is:

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

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

Now, the series is given as:

role="math" localid="1658919005396" $\frac{1}{3}+\frac{1}{6}+\frac{1}{9}+\frac{1}{12}+\cdots \dots \dots .=\frac{1}{3}\sum \frac{1}{n}$

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

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

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

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

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

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

Hence, the series is $\sum _{\mathrm{n}=2}^8\frac{1}{{\mathrm{n}}^{\frac{3}{2}}}$ 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:

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

This is not possible.