Notes: `1/3+1/5+1/7+1/9+1/11+...` Confirm that the Integral Test can be applied to the series. Then use the Integral Test to determine the...

Friday, 6 September 2013

$\displaystyle\frac{{1}}{{3}}+\frac{{1}}{{5}}+\frac{{1}}{{7}}+\frac{{1}}{{9}}+\frac{{1}}{{11}}+\ldots$ Confirm that the Integral Test can be applied to the series. Then use the Integral Test to determine the...

$\displaystyle\frac{{1}}{{3}}+\frac{{1}}{{5}}+\frac{{1}}{{7}}+\frac{{1}}{{9}}+\frac{{1}}{{11}}+\ldots\ldots\ldots.$

The series can be written as,

$\displaystyle\frac{{1}}{{{2}\cdot{1}+{1}}}+\frac{{1}}{{{2}\cdot{2}+{1}}}+\frac{{1}}{{{2}\cdot{3}+{1}}}+\frac{{1}}{{{2}\cdot{4}+{1}}}+\frac{{1}}{{{2}\cdot{5}+{1}}}+\ldots\ldots\ldots$

Based on the above pattern we can write the series as,

$\displaystyle{\sum_{{{n}={1}}}^{\infty}}\frac{{1}}{{{2}{n}+{1}}}$

The integral test is applicable if f is positive, continuous and decreasing function on the interval $\displaystyle{\left[{k},\infty\right)}$ where $\displaystyle{k}\ge{1}$ and $\displaystyle{a}_{{n}}={f{{\left({x}\right)}}}$ . Then the series 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.

For the given series $\displaystyle{a}_{{n}}=\frac{{1}}{{{2}{n}+{1}}}$

Consider $\displaystyle{f{{\left({x}\right)}}}=\frac{{1}}{{{2}{x}+{1}}}$

Refer to the attached graph of the function. From the graph...

$\displaystyle\frac{{1}}{{3}}+\frac{{1}}{{5}}+\frac{{1}}{{7}}+\frac{{1}}{{9}}+\frac{{1}}{{11}}+\ldots\ldots\ldots.$

The series can be written as,

$\displaystyle\frac{{1}}{{{2}\cdot{1}+{1}}}+\frac{{1}}{{{2}\cdot{2}+{1}}}+\frac{{1}}{{{2}\cdot{3}+{1}}}+\frac{{1}}{{{2}\cdot{4}+{1}}}+\frac{{1}}{{{2}\cdot{5}+{1}}}+\ldots\ldots\ldots$

Based on the above pattern we can write the series as,

$\displaystyle{\sum_{{{n}={1}}}^{\infty}}\frac{{1}}{{{2}{n}+{1}}}$

For the given series $\displaystyle{a}_{{n}}=\frac{{1}}{{{2}{n}+{1}}}$

Consider $\displaystyle{f{{\left({x}\right)}}}=\frac{{1}}{{{2}{x}+{1}}}$

Refer to the attached graph of the function. From the graph we can see that the function is positive, continuous and decreasing on the interval $\displaystyle{\left[{1},\infty\right)}$

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

We can apply the integral test, as the function satisfies the conditions for the integral test.

Now let's determine whether the corresponding improper integral $\displaystyle{\int_{{1}}^{\infty}}\frac{{1}}{{{2}{x}+{1}}}{\left.{d}{x}\right.}$ converges or diverges.

$\displaystyle{\int_{{1}}^{\infty}}\frac{{1}}{{{2}{x}+{1}}}{\left.{d}{x}\right.}=\lim_{{{b}\to\infty}}{\int_{{1}}^{{b}}}\frac{{1}}{{{2}{x}+{1}}}{\left.{d}{x}\right.}$

Let's first evaluate the indefinite integral $\displaystyle\int\frac{{1}}{{{2}{x}+{1}}}{\left.{d}{x}\right.}$

Apply integral substitution: $\displaystyle{u}={2}{x}+{1}$

$\displaystyle\Rightarrow{d}{u}={2}{\left.{d}{x}\right.}$

$\displaystyle\int\frac{{1}}{{{2}{x}+{1}}}{\left.{d}{x}\right.}=\int\frac{{1}}{{u}}\frac{{{d}{u}}}{{2}}$

Take the constant out and use common integral: $\displaystyle\int\frac{{1}}{{x}}{\left.{d}{x}\right.}={\ln}{\left|{x}\right|}$

$\displaystyle=\frac{{1}}{{2}}{\ln}{\left|{u}\right|}$

Substitute back $\displaystyle{u}={2}{x}+{1}$

$\displaystyle=\frac{{1}}{{2}}{\ln}{\left|{2}{x}+{1}\right|}+{C}$ where C is a constant

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

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

$\displaystyle=\infty-\frac{{\ln{{3}}}}{{2}}$

$\displaystyle=\infty$

Since the integral $\displaystyle{\int_{{1}}^{\infty}}\frac{{1}}{{{2}{x}+{1}}}{\left.{d}{x}\right.}$ diverges, we conclude from the integral test that the series diverges.

Notes

Friday, 6 September 2013

$\displaystyle\frac{{1}}{{3}}+\frac{{1}}{{5}}+\frac{{1}}{{7}}+\frac{{1}}{{9}}+\frac{{1}}{{11}}+\ldots$ 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?

Friday, 6 September 2013

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?

$\displaystyle\frac{{1}}{{3}}+\frac{{1}}{{5}}+\frac{{1}}{{7}}+\frac{{1}}{{9}}+\frac{{1}}{{11}}+\ldots$ 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?