New Notes Blog: `sum_(n=1)^oo 3^(-n)` Confirm that the Integral Test can be applied to the series. Then use the Integral Test to determine the convergence or...

Sunday, 28 December 2014

$\displaystyle{\sum_{{{n}={1}}}^{\infty}}{{3}}^{{-{n}}}$ Confirm that the Integral Test can be applied to the series. Then use the Integral Test to determine the convergence or...

Integral test is applicable if $\displaystyle{f{}}$ is positive and decreasing function on interval $\displaystyle{\left[{k},\infty\right)}$ where $\displaystyle{a}_{{n}}={f{{\left({x}\right)}}}$ .

If the integral $\displaystyle{\int_{{k}}^{\infty}}{f{{\left({x}\right)}}}{\left.{d}{x}\right.}$ is convergent then the series $\displaystyle{\sum_{{{n}={k}}}^{\infty}}{a}_{{n}}$ is also convergent.

If the integral $\displaystyle{\int_{{k}}^{\infty}}{f{{\left({x}\right)}}}{\left.{d}{x}\right.}$ is divergent then the series $\displaystyle{\sum_{{{n}={k}}}^{\infty}}{a}_{{n}}$ is also divergent.

For the series $\displaystyle{\sum_{{{n}={1}}}^{\infty}}{{3}}^{{-{n}}}$ , we have $\displaystyle{a}_{{n}}={{3}}^{{-{n}}}$ then we may let the function:

$\displaystyle{f{{\left({x}\right)}}}={{3}}^{{-{x}}}$ which has the below graph:

As...

If the integral $\displaystyle{\int_{{k}}^{\infty}}{f{{\left({x}\right)}}}{\left.{d}{x}\right.}$ is convergent then the series $\displaystyle{\sum_{{{n}={k}}}^{\infty}}{a}_{{n}}$ is also convergent.

If the integral $\displaystyle{\int_{{k}}^{\infty}}{f{{\left({x}\right)}}}{\left.{d}{x}\right.}$ is divergent then the series $\displaystyle{\sum_{{{n}={k}}}^{\infty}}{a}_{{n}}$ is also divergent.

For the series $\displaystyle{\sum_{{{n}={1}}}^{\infty}}{{3}}^{{-{n}}}$ , we have $\displaystyle{a}_{{n}}={{3}}^{{-{n}}}$ then we may let the function:

$\displaystyle{f{{\left({x}\right)}}}={{3}}^{{-{x}}}$ which has the below graph:

As shown on the graph, $\displaystyle{f{{\left({x}\right)}}}$ is positive and decreasing on the interval $\displaystyle{\left[{1},\infty\right)}$ . This confirms that we may apply the Integral test to determine the convergence or divergence of a series as:

$\displaystyle{\int_{{1}}^{\infty}}{{3}}^{{-{x}}}{\left.{d}{x}\right.}=\lim_{{{t}-\gt\infty}}{\int_{{1}}^{{t}}}{{3}}^{{-{x}}}{\left.{d}{x}\right.}$

To determine the indefinite integral of $\displaystyle{\int_{{1}}^{{t}}}{{3}}^{{-{x}}}{\left.{d}{x}\right.}$ , we may apply u-substitution by letting: $\displaystyle{u}=-{x}$ then $\displaystyle{d}{u}=-{\left.{d}{x}\right.}$ or $\displaystyle-{1}{d}{u}={\left.{d}{x}\right.}$ .

The integral becomes:

$\displaystyle\int{{3}}^{{-{x}}}{\left.{d}{x}\right.}=\int{{3}}^{{u}}\cdot-{1}{d}{u}$

$\displaystyle=-\int{{3}}^{{u}}{d}{u}$

Apply the integration formula for an exponential function: $\displaystyle\int{{a}}^{{u}}{d}{u}=\frac{{{a}}^{{u}}}{{\ln{{\left({a}\right)}}}}+{C}$ where $\displaystyle{a}$ is a constant.

$\displaystyle-\int{{3}}^{{u}}{d}{u}=-\frac{{{3}}^{{u}}}{{\ln{{\left({2}\right)}}}}$

Plugging-in $\displaystyle{u}=-{x}$ on $\displaystyle-\frac{{{3}}^{{u}}}{{\ln{{\left({3}\right)}}}}$ , we get:

$\displaystyle{\int_{{1}}^{{t}}}{{3}}^{{-{x}}}{\left.{d}{x}\right.}=-\frac{{{3}}^{{-{x}}}}{{\ln{{\left({3}\right)}}}}{{\mid}_{{1}}^{{t}}}$

$\displaystyle=-\frac{{1}}{{{{3}}^{{x}}{\ln{{\left({3}\right)}}}}}{{\mid}_{{1}}^{{t}}}$

Applying the definite integral formula: $\displaystyle{F}{\left({x}\right)}{{\mid}_{{a}}^{{b}}}={F}{\left({b}\right)}-{F}{\left({a}\right)}$ .

$\displaystyle-\frac{{1}}{{{{3}}^{{x}}{\ln{{\left({3}\right)}}}}}{{\mid}_{{1}}^{{t}}}={\left[-\frac{{1}}{{{{3}}^{{t}}{\ln{{\left({3}\right)}}}}}\right]}-{\left[-\frac{{1}}{{{{3}}^{{1}}{\ln{{\left({3}\right)}}}}}\right]}$

$\displaystyle=-\frac{{1}}{{{{3}}^{{t}}{\ln{{\left({3}\right)}}}}}+\frac{{1}}{{{3}{\ln{{\left({3}\right)}}}}}$

$\displaystyle=-\frac{{1}}{{{{3}}^{{t}}{\ln{{\left({3}\right)}}}}}+\frac{{1}}{{\ln{{\left({27}\right)}}}}$

Note: $\displaystyle{3}{\ln{{\left({3}\right)}}}={\ln{{\left({{3}}^{{3}}\right)}}}={\ln{{\left({27}\right)}}}$

Apply $\displaystyle{\int_{{1}}^{{t}}}{{3}}^{{-{x}}}{\left.{d}{x}\right.}=-\frac{{1}}{{{{3}}^{{t}}{\ln{{\left({3}\right)}}}}}+\frac{{1}}{{\ln{{\left({27}\right)}}}}$ , we get:

$\displaystyle\lim_{{{t}-\gt\infty}}{\int_{{1}}^{{t}}}{{3}}^{{-{x}}}{\left.{d}{x}\right.}=\lim_{{{t}-\gt\infty}}{\left[-\frac{{1}}{{{{3}}^{{t}}{\ln{{\left({3}\right)}}}}}+\frac{{1}}{{\ln{{\left({27}\right)}}}}\right]}$

$\displaystyle=\lim_{{{t}-\gt\infty}}-\frac{{1}}{{{{3}}^{{t}}{\ln{{\left({3}\right)}}}}}+\lim_{{{t}-\gt\infty}}\frac{{1}}{{\ln{{\left({27}\right)}}}}$

$\displaystyle={0}+\frac{{1}}{{\ln{{\left({27}\right)}}}}$

$\displaystyle=\frac{{1}}{{\ln{{\left({27}\right)}}}}$

Note: $\displaystyle{{3}}^{\infty}{\ln{{\left({3}\right)}}}=\infty$ then $\displaystyle\frac{{1}}{\infty}={0}$ .

The $\displaystyle\lim_{{{t}-\gt\infty}}{\int_{{1}}^{{t}}}{{3}}^{{-{x}}}{\left.{d}{x}\right.}=\frac{{1}}{{\ln{{\left({27}\right)}}}}$ implies the integral converges.

Conclusion:

The integral $\displaystyle{\int_{{1}}^{\infty}}{{3}}^{{-{x}}}{\left.{d}{x}\right.}$ is convergent therefore the series $\displaystyle{\sum_{{{n}={1}}}^{\infty}}{{3}}^{{-{n}}}$ must also be convergent.

New Notes Blog

Sunday, 28 December 2014

$\displaystyle{\sum_{{{n}={1}}}^{\infty}}{{3}}^{{-{n}}}$ Confirm that the Integral Test can be applied to the series. Then use the Integral Test to determine the convergence or...

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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?

Sunday, 28 December 2014

Confirm that the Integral Test can be applied to the series. Then use the Integral Test to determine the convergence or...

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?

$\displaystyle{\sum_{{{n}={1}}}^{\infty}}{{3}}^{{-{n}}}$ Confirm that the Integral Test can be applied to the series. Then use the Integral Test to determine the convergence or...

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