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

Sunday, 21 August 2016

$\displaystyle\frac{{\ln{{2}}}}{{2}}+\frac{{\ln{{3}}}}{{3}}+\frac{{\ln{{4}}}}{{4}}+\frac{{\ln{{5}}}}{{5}}+\frac{{\ln{{6}}}}{{6}}+\ldots$ Confirm that the Integral Test can be applied to the series. Then use the Integral Test to determine the...

Integral test is applicable if f is positive, continuous and decreasing function on infinite interval $\displaystyle{\left[{k},\infty\right)}$ where $\displaystyle{k}\ge{1}$ and $\displaystyle{f{{\left({n}\right)}}}={a}_{{n}}$ . Then the series $\displaystyle{\sum_{{{n}={k}}}^{\infty}}{a}_{{n}}$ converges or diverges if and only if the improper integral $\displaystyle{\int_{{k}}^{\infty}}{f{{\left({x}\right)}}}{\left.{d}{x}\right.}$ converges or diverges.

Given series is $\displaystyle\frac{{\ln{{2}}}}{{2}}+\frac{{\ln{{3}}}}{{3}}+\frac{{\ln{{4}}}}{{4}}+\frac{{\ln{{5}}}}{{5}}+\frac{{\ln{{6}}}}{{6}}+\ldots\ldots..$

The series can be written as $\displaystyle{\sum_{{{n}={1}}}^{\infty}}\frac{{\ln{{\left({n}+{1}\right)}}}}{{{n}+{1}}}$

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

Refer the attached graph for f(x),

From the graph, we observe that the function is positive, continuous and decreasing for $\displaystyle{x}\ge{2}$

We can also determine whether f(x)...

Given series is $\displaystyle\frac{{\ln{{2}}}}{{2}}+\frac{{\ln{{3}}}}{{3}}+\frac{{\ln{{4}}}}{{4}}+\frac{{\ln{{5}}}}{{5}}+\frac{{\ln{{6}}}}{{6}}+\ldots\ldots..$

The series can be written as $\displaystyle{\sum_{{{n}={1}}}^{\infty}}\frac{{\ln{{\left({n}+{1}\right)}}}}{{{n}+{1}}}$

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

Refer the attached graph for f(x),

From the graph, we observe that the function is positive, continuous and decreasing for $\displaystyle{x}\ge{2}$

We can also determine whether f(x) is decreasing by finding the derivative f'(x), such that $\displaystyle{f{'}}{\left({x}\right)}<{0}$ for $\displaystyle{x}>{1}$

Since the function satisfies the conditions for the integral test , we can apply the same.

Now let's determine the convergence or divergence of the integral $\displaystyle{\int_{{1}}^{\infty}}\frac{{\ln{{\left({x}+{1}\right)}}}}{{{x}+{1}}}{\left.{d}{x}\right.}$

$\displaystyle{\int_{{1}}^{\infty}}\frac{{\ln{{\left({x}+{1}\right)}}}}{{{x}+{1}}}{\left.{d}{x}\right.}=\lim_{{{b}\to\infty}}{\int_{{1}}^{{b}}}\frac{{\ln{{\left({x}+{1}\right)}}}}{{{x}+{1}}}{\left.{d}{x}\right.}$

Let's first evaluate the indefinite integral, $\displaystyle\int\frac{{\ln{{\left({x}+{1}\right)}}}}{{{x}+{1}}}{\left.{d}{x}\right.}$

Apply integral substitution: $\displaystyle{u}={\ln{{\left({x}+{1}\right)}}}$

$\displaystyle\Rightarrow{d}{u}=\frac{{1}}{{{x}+{1}}}{\left.{d}{x}\right.}$

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

$\displaystyle=\frac{{{u}}^{{2}}}{{2}}+{C}$ , where C is a constant

Substitute back $\displaystyle{u}={\ln{{\left({x}+{1}\right)}}}$

$\displaystyle=\frac{{1}}{{2}}{{\left[{\ln{{\left({x}+{1}\right)}}}\right]}}^{{2}}+{C}$

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

$\displaystyle=\lim_{{{b}\to\infty}}{\left[\frac{{1}}{{2}}{{\left({\ln{{\left({b}+{1}\right)}}}\right)}}^{{2}}\right]}-{\left[\frac{{1}}{{2}}{{\left({\ln{{\left({2}\right)}}}\right)}}^{{2}}\right]}$

$\displaystyle\lim_{{{x}\to\infty}}{\left({x}+{1}\right)}=\infty$

$\displaystyle\lim_{{{u}\to\infty}}{\ln{{\left({u}\right)}}}=\infty$

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

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

$\displaystyle=\infty$

Since the integral $\displaystyle{\int_{{1}}^{\infty}}\frac{{\ln{{\left({x}+{1}\right)}}}}{{{x}+{1}}}{\left.{d}{x}\right.}$ diverges, so the given series also diverges as per the integral test.

Notes

Sunday, 21 August 2016

$\displaystyle\frac{{\ln{{2}}}}{{2}}+\frac{{\ln{{3}}}}{{3}}+\frac{{\ln{{4}}}}{{4}}+\frac{{\ln{{5}}}}{{5}}+\frac{{\ln{{6}}}}{{6}}+\ldots$ Confirm that the Integral Test can be applied to the series. Then use the Integral Test to determine the...

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

Sunday, 21 August 2016

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

No comments:

Post a Comment

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

$\displaystyle\frac{{\ln{{2}}}}{{2}}+\frac{{\ln{{3}}}}{{3}}+\frac{{\ln{{4}}}}{{4}}+\frac{{\ln{{5}}}}{{5}}+\frac{{\ln{{6}}}}{{6}}+\ldots$ Confirm that the Integral Test can be applied to the series. Then use the Integral Test to determine the...

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