New Notes Blog: `sum_(n=2)^oo 1/(n(lnn)^3)` Determine the convergence or divergence of the series.

Monday, 15 December 2014

$\displaystyle{\sum_{{{n}={2}}}^{\infty}}\frac{{1}}{{{n}{{\left({\ln{{n}}}\right)}}^{{3}}}}$ Determine the convergence or divergence of the series.

The Integral test is applicable if f is positive and a 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 $\displaystyle{\sum_{{{n}={2}}}^{\infty}}\frac{{1}}{{{n}{{\left({\ln{{\left({n}\right)}}}\right)}}^{{3}}}}$ , then $\displaystyle{a}_{{n}}=\frac{{1}}{{{n}{{\left({\ln{{\left({n}\right)}}}\right)}}^{{3}}}}$ .

Then applying $\displaystyle{a}_{{n}}={f{{\left({x}\right)}}}$ , we consider: $\displaystyle{f{{\left({x}\right)}}}=\frac{{1}}{{{x}{{\left({\ln{{\left({x}\right)}}}\right)}}^{{3}}}}$ .

The graph of f(x) is:

As shown on the graph above, the function $\displaystyle{f{{\left({x}\right)}}}$ is positive and decreasing on the finite interval $\displaystyle{\left[{2},\infty\right)}$ . This implies 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:

$\displaystyle{\int_{{2}}^{\infty}}\frac{{1}}{{{x}{{\left({\ln{{\left({x}\right)}}}\right)}}^{{3}}}}=\lim_{{{t}-\gt\infty}}{\int_{{2}}^{{t}}}\frac{{1}}{{{x}{{\left({\ln{{\left({x}\right)}}}\right)}}^{{3}}}}{\left.{d}{x}\right.}$

To determine the indefinite integral of $\displaystyle{\int_{{2}}^{{t}}}\frac{{1}}{{{x}{{\left({\ln{{\left({x}\right)}}}\right)}}^{{3}}}}{\left.{d}{x}\right.}$ , we may apply u-substitution by letting:

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

The integral becomes:

$\displaystyle\int\frac{{1}}{{{x}{{\left({\ln{{\left({x}\right)}}}\right)}}^{{3}}}}{\left.{d}{x}\right.}=\int\frac{{1}}{{{\left({\ln{{\left({x}\right)}}}\right)}}^{{3}}}\cdot\frac{{1}}{{x}}{\left.{d}{x}\right.}$

$\displaystyle=\int\frac{{1}}{{{u}}^{{3}}}{d}{u}$

Apply Law of exponent: $\displaystyle\frac{{1}}{{{x}}^{{m}}}={{x}}^{{-{m}}}$ .

$\displaystyle\int\frac{{1}}{{{u}}^{{3}}}{d}{u}=\int{{u}}^{{-{3}}}{d}{u}$

Apply Power rule for integration: $\displaystyle\int{{x}}^{{n}}{\left.{d}{x}\right.}=\frac{{{x}}^{{{n}+{1}}}}{{{n}+{1}}}$ .

$\displaystyle\int{{u}}^{{-{3}}}{d}{u}=\frac{{{u}}^{{-{3}+{1}}}}{{-{3}+{1}}}$

$\displaystyle=\frac{{{u}}^{{-{2}}}}{{-{2}}}$

$\displaystyle=-\frac{{1}}{{{2}{{u}}^{{2}}}}$

Plug-in $\displaystyle{u}={\ln{{\left({x}\right)}}}$ on $\displaystyle-\frac{{1}}{{{2}{{u}}^{{2}}}}$ , we get:

$\displaystyle{\int_{{2}}^{{t}}}\frac{{1}}{{{x}{{\left({\ln{{\left({x}\right)}}}\right)}}^{{3}}}}{\left.{d}{x}\right.}=-\frac{{1}}{{{2}{{\left({\ln{{\left({x}\right)}}}\right)}}^{{2}}}}{{\mid}_{{2}}^{{t}}}$

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

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

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

Applying $\displaystyle{\int_{{1}}^{{t}}}\frac{{1}}{{{x}{{\left({\ln{{\left({x}\right)}}}\right)}}^{{3}}}}{\left.{d}{x}\right.}=-\frac{{1}}{{{2}{{\left({\ln{{\left({t}\right)}}}\right)}}^{{2}}}}+\frac{{1}}{{{2}{{\left({\ln{{\left({2}\right)}}}\right)}}^{{2}}}}$ , we get:

$\displaystyle\lim_{{{t}-\gt\infty}}{\int_{{1}}^{{t}}}\frac{{1}}{{{x}{{\left({\ln{{\left({x}\right)}}}\right)}}^{{3}}}}{\left.{d}{x}\right.}=\lim_{{{t}-\gt\infty}}{\left[-\frac{{1}}{{{2}{{\left({\ln{{\left({t}\right)}}}\right)}}^{{2}}}}+\frac{{1}}{{{2}{{\left({\ln{{\left({2}\right)}}}\right)}}^{{2}}}}\right]}$

$\displaystyle={0}+\frac{{1}}{{{2}{{\left({\ln{{\left({2}\right)}}}\right)}}^{{2}}}}$

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

Note: $\displaystyle\lim_{{{t}-\gt\infty}}\frac{{1}}{{{2}{{\left({\ln{{\left({2}\right)}}}\right)}}^{{2}}}}=\frac{{1}}{{{2}{{\left({\ln{{\left({2}\right)}}}\right)}}^{{2}}}}$ and

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

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

$\displaystyle=-{0}{\quad\text{or}\quad}{0}$

The $\displaystyle\lim_{{{t}-\gt\infty}}{\int_{{2}}^{{t}}}\frac{{1}}{{{x}{{\left({\ln{{\left({x}\right)}}}\right)}}^{{3}}}}{\left.{d}{x}\right.}=\frac{{1}}{{{2}{{\left({\ln{{\left({2}\right)}}}\right)}}^{{2}}}}$ implies that the integral converges.

Conclusion: The integral $\displaystyle{\int_{{2}}^{\infty}}\frac{{1}}{{{x}{{\left({\ln{{\left({x}\right)}}}\right)}}^{{3}}}}$ is convergent therefore the series $\displaystyle{\sum_{{{n}={2}}}^{\infty}}\frac{{1}}{{{n}{{\left({\ln{{\left({n}\right)}}}\right)}}^{{3}}}}$ must also be convergent.

New Notes Blog

Monday, 15 December 2014

$\displaystyle{\sum_{{{n}={2}}}^{\infty}}\frac{{1}}{{{n}{{\left({\ln{{n}}}\right)}}^{{3}}}}$ Determine the convergence or divergence of the series.

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In "By the Waters of Babylon," under the leadership of John, what do you think the Hill People will do with their society?

Monday, 15 December 2014

Determine the convergence or divergence of the series.

No comments:

Post a Comment

In &quot;By the Waters of Babylon,&quot; under the leadership of John, what do you think the Hill People will do with their society?

$\displaystyle{\sum_{{{n}={2}}}^{\infty}}\frac{{1}}{{{n}{{\left({\ln{{n}}}\right)}}^{{3}}}}$ Determine the convergence or divergence of the series.

In "By the Waters of Babylon," under the leadership of John, what do you think the Hill People will do with their society?