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

Friday, 11 December 2015

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

$\displaystyle{\sum_{{{n}={1}}}^{\infty}}\frac{{\arctan{{\left({n}\right)}}}}{{{{n}}^{{2}}+{1}}}$

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 $\displaystyle{a}_{{n}}=\frac{{\arctan{{\left({n}\right)}}}}{{{{n}}^{{2}}+{1}}}$

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

Refer the attached graph of the function. From the graph, we observe that the function is positive , continuous and decreasing for $\displaystyle{x}\ge{1}$

We an apply integral test as the function...

$\displaystyle{\sum_{{{n}={1}}}^{\infty}}\frac{{\arctan{{\left({n}\right)}}}}{{{{n}}^{{2}}+{1}}}$

For the given series $\displaystyle{a}_{{n}}=\frac{{\arctan{{\left({n}\right)}}}}{{{{n}}^{{2}}+{1}}}$

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

Refer the attached graph of the function. From the graph, we observe that the function is positive , continuous and decreasing for $\displaystyle{x}\ge{1}$

We an apply integral test as the function satisfies all the conditions for the integral test.

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

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

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

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

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

$\displaystyle=\int{u}{d}{u}$

Apply power rule,

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

Substitute back $\displaystyle{u}={\arctan{{\left({x}\right)}}}$

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

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

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

$\displaystyle=\frac{{1}}{{2}}{\left[{{\left(\frac{\pi}{{2}}\right)}}^{{2}}-{{\left(\frac{\pi}{{4}}\right)}}^{{2}}\right]}$

$\displaystyle=\frac{{1}}{{2}}{\left(\frac{{\pi}^{{2}}}{{4}}-\frac{{\pi}^{{2}}}{{16}}\right)}$

$\displaystyle=\frac{{1}}{{2}}{\left(\frac{{{4}{\pi}^{{2}}-{\pi}^{{2}}}}{{16}}\right)}$

$\displaystyle=\frac{{3}}{{32}}{\pi}^{{2}}$

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

Notes

Friday, 11 December 2015

$\displaystyle{\sum_{{{n}={1}}}^{\infty}}\frac{{\arctan{{\left({n}\right)}}}}{{{{n}}^{{2}}+{1}}}$ 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, 11 December 2015

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?

$\displaystyle{\sum_{{{n}={1}}}^{\infty}}\frac{{\arctan{{\left({n}\right)}}}}{{{{n}}^{{2}}+{1}}}$ 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?