Notes: `sum_(n=1)^oo n/(n^4+1)` Confirm that the Integral Test can be applied to the series. Then use the Integral Test to determine the convergence...

Thursday, 17 July 2014

$\displaystyle{\sum_{{{n}={1}}}^{\infty}}\frac{{n}}{{{{n}}^{{4}}+{1}}}$ Confirm that the Integral Test can be applied to the series. Then use the Integral Test to determine the convergence...

$\displaystyle{\sum_{{{n}={1}}}^{\infty}}\frac{{n}}{{{{n}}^{{4}}+{1}}}$

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 $\displaystyle{a}_{{n}}={f{{\left({x}\right)}}}$ . Then the series $\displaystyle{\sum_{{{n}={1}}}^{\infty}}{a}_{{n}}$ converges or diverges if and only if the improper integral $\displaystyle{\int_{{1}}^{\infty}}{f{{\left({x}\right)}}}{\left.{d}{x}\right.}$ converges or diverges.

For the given series $\displaystyle{a}_{{n}}=\frac{{n}}{{{{n}}^{{4}}+{1}}}$

Consider $\displaystyle{f{{\left({x}\right)}}}=\frac{{x}}{{{{x}}^{{4}}+{1}}}$

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

We can determine...

$\displaystyle{\sum_{{{n}={1}}}^{\infty}}\frac{{n}}{{{{n}}^{{4}}+{1}}}$

For the given series $\displaystyle{a}_{{n}}=\frac{{n}}{{{{n}}^{{4}}+{1}}}$

Consider $\displaystyle{f{{\left({x}\right)}}}=\frac{{x}}{{{{x}}^{{4}}+{1}}}$

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

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

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

Now let's determine whether the corresponding improper integral $\displaystyle{\int_{{1}}^{\infty}}\frac{{x}}{{{{x}}^{{4}}+{1}}}{\left.{d}{x}\right.}$ converges or diverges.

$\displaystyle{\int_{{1}}^{\infty}}\frac{{x}}{{{{x}}^{{4}}+{1}}}{\left.{d}{x}\right.}=\lim_{{{b}\to\infty}}{\int_{{1}}^{{b}}}\frac{{x}}{{{{x}}^{{4}}+{1}}}{\left.{d}{x}\right.}$

Let's first evaluate the indefinite integral $\displaystyle\int\frac{{x}}{{{{x}}^{{4}}+{1}}}{\left.{d}{x}\right.}$

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

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

$\displaystyle\int\frac{{x}}{{{{x}}^{{4}}+{1}}}{\left.{d}{x}\right.}=\int\frac{{1}}{{{{u}}^{{2}}+{1}}}\frac{{{d}{u}}}{{2}}$

Take the constant out and use common integral: $\displaystyle\int\frac{{1}}{{{{x}}^{{2}}+{1}}}{\left.{d}{x}\right.}={\arctan{{\left({x}\right)}}}+{C}$

$\displaystyle=\frac{{1}}{{2}}{\arctan{{\left({u}\right)}}}+{C}$

Substitute back $\displaystyle{u}={{x}}^{{2}}$

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

$\displaystyle{\int_{{1}}^{\infty}}\frac{{x}}{{{{x}}^{{4}}+{1}}}{\left.{d}{x}\right.}=\lim_{{{b}\to\infty}}{{\left[\frac{{1}}{{2}}{\arctan{{\left({{x}}^{{2}}\right)}}}\right]}_{{1}}^{{b}}}$

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

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

$\displaystyle=\lim_{{{b}\to\infty}}{\left[\frac{{1}}{{2}}{\arctan{{\left({{b}}^{{2}}\right)}}}\right]}-{\left[\frac{{1}}{{2}}{\left(\frac{\pi}{{4}}\right)}\right]}$

$\displaystyle=\lim_{{{b}\to\infty}}{\left[\frac{{1}}{{2}}{\arctan{{\left({{b}}^{{2}}\right)}}}\right]}-\frac{\pi}{{8}}$

$\displaystyle=\frac{{1}}{{2}}\lim_{{{b}\to\infty}}{\arctan{{\left({{b}}^{{2}}\right)}}}-\frac{\pi}{{8}}$

Now $\displaystyle\lim_{{{b}\to\infty}}{\left({{b}}^{{2}}\right)}=\infty$

$\displaystyle=\frac{{1}}{{2}}{\left(\frac{\pi}{{2}}\right)}-\frac{\pi}{{8}}$ [by applying the common limit: $\displaystyle\lim_{{{u}\to\infty}}{\arctan{{\left({u}\right)}}}=\frac{\pi}{{2}}$ ]

$\displaystyle=\frac{\pi}{{4}}-\frac{\pi}{{8}}$

$\displaystyle=\frac{\pi}{{8}}$

Since the integral $\displaystyle{\int_{{1}}^{\infty}}\frac{{x}}{{{{x}}^{{4}}+{1}}}{\left.{d}{x}\right.}$ converges, we conclude from the integral test that the series converges.

Notes

Thursday, 17 July 2014

$\displaystyle{\sum_{{{n}={1}}}^{\infty}}\frac{{n}}{{{{n}}^{{4}}+{1}}}$ 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?

Thursday, 17 July 2014

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?

$\displaystyle{\sum_{{{n}={1}}}^{\infty}}\frac{{n}}{{{{n}}^{{4}}+{1}}}$ 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?