New Notes Blog: `sum_(n=1)^oo n/(n^4+2n^2+1)` Confirm that the Integral Test can be applied to the series. Then use the Integral Test to determine the...

Wednesday, 28 January 2015

$\displaystyle{\sum_{{{n}={1}}}^{\infty}}\frac{{n}}{{{{n}}^{{4}}+{2}{{n}}^{{2}}+{1}}}$ Confirm that the Integral Test can be applied to the series. Then use the Integral Test to determine the...

$\displaystyle{\sum_{{{n}={1}}}^{\infty}}\frac{{n}}{{{{n}}^{{4}}+{2}{{n}}^{{2}}+{1}}}$

The integral test is applicable if f is positive, continuous and decreasing function on the infinite 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 $\displaystyle{a}_{{n}}=\frac{{n}}{{{{n}}^{{4}}+{2}{{n}}^{{2}}+{1}}}$

Consider $\displaystyle{f{{\left({x}\right)}}}=\frac{{x}}{{{{x}}^{{4}}+{2}{{x}}^{{2}}+{1}}}$

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

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

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

$\displaystyle{\sum_{{{n}={1}}}^{\infty}}\frac{{n}}{{{{n}}^{{4}}+{2}{{n}}^{{2}}+{1}}}$

For the given series $\displaystyle{a}_{{n}}=\frac{{n}}{{{{n}}^{{4}}+{2}{{n}}^{{2}}+{1}}}$

Consider $\displaystyle{f{{\left({x}\right)}}}=\frac{{x}}{{{{x}}^{{4}}+{2}{{x}}^{{2}}+{1}}}$

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

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

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

Apply the quotient rule to find the derivative,

$\displaystyle{f{'}}{\left({x}\right)}=\frac{{{{\left({{x}}^{{2}}+{1}\right)}}^{{2}}\frac{{d}}{{\left.{d}{x}}}{\left({x}\right)}-{x}\frac{{d}}{{\left.{d}{x}}}{{\left({{x}}^{{2}}+{1}\right)}}^{{2}}}}{{{\left({{x}}^{{2}}+{1}\right)}}^{{4}}}$

$\displaystyle{f{'}}{\left({x}\right)}=\frac{{{{\left({{x}}^{{2}}+{1}\right)}}^{{2}}-{x}{\left({2}{\left({{x}}^{{2}}+{1}\right)}{2}{x}\right)}}}{{{\left({{x}}^{{2}}+{1}\right)}}^{{4}}}$

$\displaystyle{f{'}}{\left({x}\right)}=\frac{{{\left({{x}}^{{2}}+{1}\right)}{\left({{x}}^{{2}}+{1}-{4}{{x}}^{{2}}\right)}}}{{{\left({{x}}^{{2}}+{1}\right)}}^{{4}}}$

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

$\displaystyle{f{'}}{\left({x}\right)}=-\frac{{{3}{{x}}^{{2}}-{1}}}{{{\left({{x}}^{{2}}+{1}\right)}}^{{3}}}<{0}$

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

Now let's determine the convergence or divergence of the improper integral as follows:

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

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

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

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

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

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

Apply the power rule,

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

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

Substitute back $\displaystyle{u}={\left({{x}}^{{2}}+{1}\right)}$

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

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

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

$\displaystyle=-\frac{{1}}{{2}}{\left[{0}-\frac{{1}}{{2}}\right]}$

$\displaystyle=\frac{{1}}{{4}}$

Since the integral $\displaystyle{\int_{{1}}^{\infty}}\frac{{x}}{{{{x}}^{{4}}+{2}{{x}}^{{2}}+{1}}}{\left.{d}{x}\right.}$ converges, we conclude from the integral test that the series $\displaystyle{\sum_{{{n}={1}}}^{\infty}}\frac{{n}}{{{{n}}^{{4}}+{2}{{n}}^{{2}}+{1}}}$ converges.

New Notes Blog

Wednesday, 28 January 2015

$\displaystyle{\sum_{{{n}={1}}}^{\infty}}\frac{{n}}{{{{n}}^{{4}}+{2}{{n}}^{{2}}+{1}}}$ Confirm that the Integral Test can be applied to the series. Then use the Integral Test to determine the...

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In "By the Waters of Babylon," under the leadership of John, what do you think the Hill People will do with their society?

Wednesday, 28 January 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

In &quot;By the Waters of Babylon,&quot; under the leadership of John, what do you think the Hill People will do with their society?

$\displaystyle{\sum_{{{n}={1}}}^{\infty}}\frac{{n}}{{{{n}}^{{4}}+{2}{{n}}^{{2}}+{1}}}$ Confirm that the Integral Test can be applied to the series. Then use the Integral Test to determine the...

In "By the Waters of Babylon," under the leadership of John, what do you think the Hill People will do with their society?