Notes: `sum_(n=1)^oo ln(n)/n^2` Confirm that the Integral Test can be applied to the series. Then use the Integral Test to determine the convergence...

The Integral test is applicable if $\displaystyle{f{}}$ is positive and decreasing function on infinite interval $\displaystyle{\left[{k},\infty\right)}$ where $\displaystyle{k}\geq{1}$ 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.

For the given series $\displaystyle{\sum_{{{n}={1}}}^{\infty}}\frac{{\ln{{\left({n}\right)}}}}{{{n}}^{{2}}}$ , the $\displaystyle{a}_{{n}}=\frac{{\ln{{\left({n}\right)}}}}{{{n}}^{{2}}}$ .

Then applying $\displaystyle{a}_{{n}}={f{{\left({x}\right)}}}$ , we consider:

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

The graph of f(x) is:

As shown on the graph, f is positive on the finite interval $\displaystyle{\left[{1},\infty\right)}$ . To verify of 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 $\displaystyle{u}'=\frac{{1}}{{x}}$

$\displaystyle{v}={{x}}^{{2}}$ then $\displaystyle{v}'={2}{x}$

Applying the formula,we get:

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

$\displaystyle=\frac{{{x}-{2}{x}{\ln{{\left({x}\right)}}}}}{{{x}}^{{4}}}$

$\displaystyle=\frac{{{1}-{2}{\ln{{\left({x}\right)}}}}}{{{x}}^{{3}}}$

Note that $\displaystyle{1}-{2}{\ln{{\left({x}\right)}}}\lt{0}$ for larger values of x which means $\displaystyle{f{'}}{\left({x}\right)}\lt{0}$ .Based on the First derivative test, if $\displaystyle{f{'}}{\left({x}\right)}$ has a negative value then the function $\displaystyle{f{{\left({x}\right)}}}$ is decreasing for a given interval $\displaystyle{I}$ . This confirms that the function is ultimately decreasing as $\displaystyle{x}-\gt\infty$ . Therefore, we may apply the Integral test to confirm the convergence or divergence of the given series.

We may determine the convergence or divergence of the improper integral as:

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

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

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

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

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

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

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

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

$\displaystyle=-\frac{{1}}{{x}}$

The integral becomes:

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

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

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

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

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

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

$\displaystyle-\frac{{\ln{{\left({x}\right)}}}}{{x}}-\frac{{1}}{{x}}{{\mid}_{{1}}^{{t}}}={\left[-\frac{{\ln{{\left({t}\right)}}}}{{t}}-\frac{{1}}{{t}}\right]}-{\left[-\frac{{\ln{{\left({1}\right)}}}}{{1}}-\frac{{1}}{{1}}\right]}$

$\displaystyle={\left[-\frac{{\ln{{\left({t}\right)}}}}{{t}}-\frac{{1}}{{t}}\right]}-{\left[-{0}-{1}\right]}$

$\displaystyle={\left[-\frac{{\ln{{\left({t}\right)}}}}{{t}}-\frac{{1}}{{t}}\right]}-{\left[-{1}\right]}$

$\displaystyle=-\frac{{\ln{{\left({t}\right)}}}}{{t}}-\frac{{1}}{{t}}+{1}$

Apply $\displaystyle{\int_{{1}}^{{t}}}\frac{{\ln{{\left({x}\right)}}}}{{{x}}^{{2}}}{\left.{d}{x}\right.}=-\frac{{\ln{{\left({t}\right)}}}}{{t}}-\frac{{1}}{{t}}+{1}$ , we get:

$\displaystyle\lim_{{{t}-\gt\infty}}{\int_{{1}}^{{t}}}\frac{{\ln{{\left({x}\right)}}}}{{{x}}^{{2}}}{\left.{d}{x}\right.}=\lim_{{{t}-\gt\infty}}{\left[-\frac{{\ln{{\left({t}\right)}}}}{{t}}-\frac{{1}}{{t}}+{1}\right]}$

$\displaystyle=-{0}-{0}+{1}$

$\displaystyle={1}$

Note: $\displaystyle\lim_{{{t}-\gt\infty}}{1}={1}$

$\displaystyle\lim_{{{t}-\gt\infty}}\frac{{1}}{{t}}=\frac{{1}}{\infty}{\quad\text{or}\quad}{0}$

$\displaystyle\lim_{{{t}-\gt\infty}}-\frac{{\ln{{\left({t}\right)}}}}{{t}}=\frac{{\lim_{{{t}-\gt\infty}}-{\ln{{\left({t}\right)}}}}}{{\lim_{{{t}-\gt\infty}}{t}}}$

$\displaystyle=-\frac{\infty}{\infty}$

Apply L' Hospitals rule:

$\displaystyle\lim_{{{t}-\gt\infty}}-\frac{{\ln{{\left({t}\right)}}}}{{t}}=\lim_{{{t}-\gt\infty}}-\frac{{\frac{{1}}{{t}}}}{{1}}$

$\displaystyle=\lim_{{{t}-\gt\infty}}-\frac{{1}}{{t}}$

$\displaystyle=-\frac{{1}}{\infty}{\quad\text{or}\quad}{0}$

The $\displaystyle\lim_{{{t}-\gt\infty}}{\int_{{1}}^{{t}}}\frac{{\ln{{\left({x}\right)}}}}{{{x}}^{{2}}}{\left.{d}{x}\right.}={1}$ implies that the integral converges.

Conclusion: The integral $\displaystyle{\int_{{1}}^{\infty}}\frac{{\ln{{\left({x}\right)}}}}{{{x}}^{{2}}}{\left.{d}{x}\right.}$ is convergent therefore the series $\displaystyle{\sum_{{{n}={1}}}^{\infty}}\frac{{\ln{{\left({n}\right)}}}}{{{n}}^{{2}}}$ must also be convergent.

Notes

Monday, 26 June 2017

$\displaystyle{\sum_{{{n}={1}}}^{\infty}}\frac{{\ln{{\left({n}\right)}}}}{{{n}}^{{2}}}$ Confirm that the Integral Test can be applied to the series. Then use the Integral Test to determine the convergence...

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How are race, gender, and class addressed in Oliver Optic's Rich and Humble?

Monday, 26 June 2017

Confirm that the Integral Test can be applied to the series. Then use the Integral Test to determine the convergence...

No comments:

Post a Comment

How are race, gender, and class addressed in Oliver Optic&#39;s Rich and Humble?

$\displaystyle{\sum_{{{n}={1}}}^{\infty}}\frac{{\ln{{\left({n}\right)}}}}{{{n}}^{{2}}}$ Confirm that the Integral Test can be applied to the series. Then use the Integral Test to determine the convergence...

How are race, gender, and class addressed in Oliver Optic's Rich and Humble?