Notes: `ln2/sqrt(2)+ln3/sqrt(3)+ln4/sqrt(4)+ln5/sqrt(5)+ln6/sqrt(6)+...` Confirm that the Integral Test can be applied to the series. Then use the...

For the series: $\displaystyle\frac{{\ln{{\left({2}\right)}}}}{\sqrt{{{2}}}}+\frac{{\ln{{\left({3}\right)}}}}{\sqrt{{{3}}}}+\frac{{\ln{{\left({4}\right)}}}}{\sqrt{{{4}}}}+\frac{{\ln{{\left({5}\right)}}}}{\sqrt{{{5}}}}+\frac{{\ln{{\left({6}\right)}}}}{\sqrt{{{6}}}}+\ldots$ , it follows the formula $\displaystyle{\sum_{{{n}={2}}}^{\infty}}\frac{{\ln{{\left({n}\right)}}}}{\sqrt{{{n}}}}$ where $\displaystyle{a}_{{n}}=\frac{{\ln{{\left({n}\right)}}}}{\sqrt{{{n}}}}$ . To confirm if the Integral test will be applicable, we let $\displaystyle{f{{\left({x}\right)}}}=\frac{{\ln{{\left({x}\right)}}}}{\sqrt{{{x}}}}$ .

Graph of the function $\displaystyle{f{{\left({x}\right)}}}$ :

Maximize view:

As shown on the graphs, $\displaystyle{f{}}$ is positive and continuous on the finite interval $\displaystyle{\left[{1},\infty\right)}$ . To verify if the function will eventually decreases on the given interval, we may consider derivative of the function.

Apply Quotient rule for derivative: $\displaystyle\frac{{d}}{{\left.{d}{x}}}{\left(\frac{{u}}{{v}}\right)}=\frac{{{u}'\cdot{v}-{v}'\cdot{u}}}{{{v}}^{{2}}}$ .

Let $\displaystyle{u}={\ln{{\left({x}\right)}}}$ then $\displaystyle{u}'=\frac{{1}}{{x}}$

$\displaystyle{v}=\sqrt{{{x}}}$ or $\displaystyle{{x}}^{{\frac{{1}}{{2}}}}$ then $\displaystyle{v}'=\frac{{1}}{{{2}\sqrt{{{x}}}}}$

Applying the Quotient rule, we get:

$\displaystyle{f{'}}{\left({x}\right)}=\frac{{\frac{{1}}{{x}}\cdot\sqrt{{{x}}}-\frac{{1}}{{{2}\sqrt{{{x}}}}}\cdot{\ln{{\left({x}\right)}}}}}{{{\left(\sqrt{{{x}}}\right)}}^{{2}}}$

$\displaystyle=\frac{{\frac{{1}}{\sqrt{{{x}}}}-\frac{{\ln{{\left({x}\right)}}}}{{{2}\sqrt{{{x}}}}}}}{{x}}$

$\displaystyle=\frac{{\frac{{{2}-{\ln{{\left({x}\right)}}}}}{\sqrt{{{x}}}}}}{{x}}$

$\displaystyle={\left(\frac{{{2}-{\ln{{\left({x}\right)}}}}}{\sqrt{{{x}}}}\right)}\cdot\frac{{1}}{{x}}$

$\displaystyle=\frac{{{2}-{\ln{{\left({x}\right)}}}}}{{{x}\sqrt{{{x}}}}}$

or $\displaystyle\frac{{{2}-{\ln{{\left({x}\right)}}}}}{{{x}}^{{\frac{{3}}{{2}}}}}$

Note that $\displaystyle{2}-{\ln{{\left({x}\right)}}}\lt{0}$ for higher values of x which means $\displaystyle{f{'}}{\left({x}\right)}\lt{0}$ .

Aside from this, we may verify by solving critical values of x .

Apply First derivative test: f'(c) =0 such that x =c as critical values.

$\displaystyle\frac{{{2}-{\ln{{\left({x}\right)}}}}}{{{x}}^{{\frac{{3}}{{2}}}}}={0}$

$\displaystyle{2}-{\ln{{\left({x}\right)}}}={0}$

$\displaystyle{\ln{{\left({x}\right)}}}={2}$

$\displaystyle{x}={{e}}^{{2}}$

$\displaystyle{x}\approx{7.389}$

Using $\displaystyle{f{'}}{\left({7}\right)}\approx{0.0015}$ , it satisfy $\displaystyle{f{'}}{\left({x}\right)}\gt{0}$ therefore the function is increasing on the left side of $\displaystyle{x}={{e}}^{{2}}$ .

Using $\displaystyle{f{'}}{\left({8}\right)}\approx-{0.0018}$ , it satisfy $\displaystyle{f{'}}{\left({x}\right)}\lt{0}$ therefore the function is decreasing on the right side of $\displaystyle{x}={{e}}^{{2}}$ .

Then, we may conclude that the function $\displaystyle{f{{\left({x}\right)}}}$ is decreasing for an interval $\displaystyle{\left[{8},\infty\right)}$ .

This confirms that the function is ultimately positive, continuous, and decreasing for an interval $\displaystyle{\left[{8},\infty\right)}$ . Therefore, we may apply the Integral test.

Note: Integral test is applicable if f is positive, continuous , and decreasing function on interval $\displaystyle{\left[{k},\infty\right)}$ and $\displaystyle{a}_{{n}}={f{{\left({x}\right)}}}$ . Then the series $\displaystyle{\sum_{{{n}={k}}}^{\infty}}{a}_{{n}}$ converges if and only if the improper integral $\displaystyle{\int_{{k}}^{\infty}}{f{{\left({x}\right)}}}{\left.{d}{x}\right.}$ converges. If the integral diverges then the series also diverges.

To determine the convergence or divergence of the given series, we may apply improper integral as:

$\displaystyle{\int_{{8}}^{\infty}}\frac{{\ln{{\left({x}\right)}}}}{\sqrt{{{x}}}}{\left.{d}{x}\right.}=\lim_{{{t}-\gt\infty}}{\int_{{8}}^{{t}}}\frac{{\ln{{\left({x}\right)}}}}{\sqrt{{{x}}}}{\left.{d}{x}\right.}$

or $\displaystyle\lim_{{{t}-\gt\infty}}{\int_{{8}}^{{t}}}\frac{{\ln{{\left({x}\right)}}}}{{{x}}^{{\frac{{1}}{{2}}}}}{\left.{d}{x}\right.}$

To determine the indefinite integral of $\displaystyle{\int_{{8}}^{{t}}}\frac{{\ln{{\left({x}\right)}}}}{{{x}}^{{\frac{{1}}{{2}}}}}{\left.{d}{x}\right.}$ , we may apply integration by parts: $\displaystyle\int{u}{d}{v}={u}{v}-\int{v}{d}{u}$

$\displaystyle{u}={\ln{{\left({x}\right)}}}$ then $\displaystyle{d}{u}=\frac{{1}}{{x}}{\left.{d}{x}\right.}$ .

$\displaystyle{d}{v}=\frac{{1}}{{{x}}^{{\frac{{1}}{{2}}}}}{\left.{d}{x}\right.}$ then $\displaystyle{v}=\int\frac{{1}}{{{x}}^{{\frac{{1}}{{2}}}}}{\left.{d}{x}\right.}={2}\sqrt{{{x}}}$

Note: To determine v, apply Power rule for integration $\displaystyle\int{{x}}^{{n}}{\left.{d}{x}\right.}=\frac{{{x}}^{{{n}+{1}}}}{{{n}+{1}}}.$

$\displaystyle\int\frac{{1}}{{{x}}^{{\frac{{1}}{{2}}}}}{\left.{d}{x}\right.}=\int{{x}}^{{-\frac{{1}}{{2}}}}{\left.{d}{x}\right.}$

$\displaystyle=\frac{{{x}}^{{-\frac{{1}}{{2}}+{1}}}}{{-\frac{{1}}{{2}}+{1}}}$

$\displaystyle=\frac{{{x}}^{{\frac{{1}}{{2}}}}}{{\frac{{1}}{{2}}}}$

$\displaystyle={{x}}^{{\frac{{1}}{{2}}}}\cdot\frac{{2}}{{1}}$

$\displaystyle={2}{{x}}^{{\frac{{1}}{{2}}}}$ or $\displaystyle{2}\sqrt{{{x}}}$

The integral becomes:

$\displaystyle{\int_{{8}}^{{t}}}\frac{{\ln{{\left({x}\right)}}}}{\sqrt{{{x}}}}{\left.{d}{x}\right.}={\ln{{\left({x}\right)}}}\cdot{2}\sqrt{{{x}}}-\int{2}\sqrt{{{x}}}\cdot\frac{{1}}{{x}}{\left.{d}{x}\right.}$

$\displaystyle={2}\sqrt{{{x}}}{\ln{{\left({x}\right)}}}-\int{2}{{x}}^{{\frac{{1}}{{2}}}}\cdot{{x}}^{{-{1}}}{\left.{d}{x}\right.}$

$\displaystyle={2}\sqrt{{{x}}}{\ln{{\left({x}\right)}}}-\int{2}{{x}}^{{-\frac{{1}}{{2}}}}{\left.{d}{x}\right.}$

$\displaystyle={2}\sqrt{{{x}}}{\ln{{\left({x}\right)}}}-{2}\int{{x}}^{{-\frac{{1}}{{2}}}}{\left.{d}{x}\right.}$

$\displaystyle={\left[{2}\sqrt{{{x}}}{\ln{{\left({x}\right)}}}-{2}{\left({2}\sqrt{{{x}}}\right)}\right]}{{\mid}_{{8}}^{{t}}}$

$\displaystyle={\left[{2}\sqrt{{{x}}}{\ln{{\left({x}\right)}}}-{4}\sqrt{{{x}}}\right]}{{\mid}_{{8}}^{{t}}}$

Apply definite integral formula: $\displaystyle{F}{\left({x}\right)}{{\mid}_{{a}}^{{b}}}={F}{\left({b}\right)}-{F}{\left({a}\right)}$ .

$\displaystyle{\left[{2}\sqrt{{{x}}}{\ln{{\left({x}\right)}}}-{4}\sqrt{{{x}}}\right]}{{\mid}_{{8}}^{{t}}}={\left[{2}\sqrt{{{t}}}{\ln{{\left({t}\right)}}}-{4}\sqrt{{{t}}}\right]}-{\left[{2}\sqrt{{{8}}}{\ln{{\left({8}\right)}}}-{4}\sqrt{{{8}}}\right]}$

$\displaystyle={2}\sqrt{{{t}}}{\ln{{\left({t}\right)}}}-{4}\sqrt{{{t}}}-{2}\sqrt{{{8}}}{\ln{{\left({8}\right)}}}+{4}\sqrt{{{8}}}$

$\displaystyle={2}\sqrt{{{t}}}{\ln{{\left({t}\right)}}}-{4}\sqrt{{{t}}}-{4}\sqrt{{{2}}}{\ln{{\left({8}\right)}}}+{8}\sqrt{{{2}}}$

Note: $\displaystyle\sqrt{{{8}}}={2}\sqrt{{{2}}}$

Applying $\displaystyle{\int_{{8}}^{{t}}}\frac{{\ln{{\left({x}\right)}}}}{\sqrt{{{x}}}}{\left.{d}{x}\right.}={2}\sqrt{{{t}}}{\ln{{\left({t}\right)}}}-{4}\sqrt{{{t}}}-{4}\sqrt{{{2}}}{\ln{{\left({8}\right)}}}+{8}\sqrt{{{2}}}$ , we get:

$\displaystyle\lim_{{{t}-\gt\infty}}{\int_{{2}}^{{t}}}\frac{{\ln{{\left({x}\right)}}}}{\sqrt{{{x}}}}{\left.{d}{x}\right.}=\lim_{{{t}-\gt\infty}}{\left[{2}\sqrt{{{t}}}{\ln{{\left({t}\right)}}}-{4}\sqrt{{{t}}}-{4}\sqrt{{{2}}}{\ln{{\left({8}\right)}}}+{8}\sqrt{{{2}}}\right]}$

$\displaystyle=\lim_{{{t}-\gt\infty}}{2}\sqrt{{{t}}}{\ln{{\left({t}\right)}}}-\lim_{{{t}-\gt\infty}}{4}\sqrt{{{t}}}-\lim_{{{t}-\gt\infty}}{4}\sqrt{{{2}}}{\ln{{\left({8}\right)}}}+\lim_{{{t}-\gt\infty}}{8}\sqrt{{{2}}}$

$\displaystyle=\infty-\infty-{4}\sqrt{{{2}}}{\ln{{\left({8}\right)}}}+{8}\sqrt{{{2}}}$

$\displaystyle=\infty$

The $\displaystyle\lim_{{{t}-\gt\infty}}{\int_{{8}}^{{t}}}\frac{{\ln{{\left({x}\right)}}}}{\sqrt{{{x}}}}{\left.{d}{x}\right.}=\infty$ implies that the integral diverges.

Conclusion:

The integral $\displaystyle{\int_{{8}}^{\infty}}\frac{{\ln{{\left({x}\right)}}}}{\sqrt{{{x}}}}{\left.{d}{x}\right.}$ is divergent therefore the series $\displaystyle{\sum_{{{n}={2}}}^{\infty}}\frac{{\ln{{\left({n}\right)}}}}{\sqrt{{{n}}}}$ must also be divergent.