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

Saturday, 9 July 2016

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

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

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 $\displaystyle{a}_{{n}}=\frac{{1}}{\sqrt{{{n}+{2}}}}$

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

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

Let's determine whether the function is decreasing...

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

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

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

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

Let's determine whether the function is decreasing by finding the derivative $\displaystyle{f{'}}{\left({x}\right)}$

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

$\displaystyle{f{'}}{\left({x}\right)}=-\frac{{1}}{{2}}{{\left({x}+{2}\right)}}^{{-\frac{{3}}{{2}}}}$

$\displaystyle{f{'}}{\left({x}\right)}=-\frac{{1}}{{{2}{{\left({x}+{2}\right)}}^{{\frac{{3}}{{2}}}}}}$

$\displaystyle{f{'}}{\left({x}\right)}<{0}$ which implies that the function is decreasing.

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

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

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

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

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

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

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

Apply the power rule,

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

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

$\displaystyle={2}\sqrt{{{u}}}$

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

$\displaystyle={2}\sqrt{{{x}+{2}}}+{C}$ where C is a constant

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

$\displaystyle=\lim_{{{b}\to\infty}}{\left[{2}\sqrt{{{b}+{2}}}-{2}\sqrt{{{1}+{2}}}\right]}$

$\displaystyle={2}\lim_{{{b}\to\infty}}\sqrt{{{b}+{2}}}-{2}\sqrt{{{3}}}$

$\displaystyle={2}{\left(\infty\right)}-{2}\sqrt{{{3}}}$

$\displaystyle=\infty-{2}\sqrt{{{3}}}$

$\displaystyle=\infty$

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

Notes

Saturday, 9 July 2016

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

Saturday, 9 July 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?

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