New Notes Blog: `sum_(n=1)^oo 5/(4^n+1)` Use the Limit Comparison Test to determine the convergence or divergence of the series.

Wednesday, 3 September 2014

$\displaystyle{\sum_{{{n}={1}}}^{\infty}}\frac{{5}}{{{{4}}^{{n}}+{1}}}$ Use the Limit Comparison Test to determine the convergence or divergence of the series.

Limit comparison test is applicable when $\displaystyle\sum{a}_{{n}}$ and $\displaystyle\sum{b}_{{n}}$ are series with positive terms. If $\displaystyle\lim_{{{n}\to\infty}}\frac{{a}_{{n}}}{{b}_{{n}}}={L}$ where L is a finite number and $\displaystyle{L}>{0}$ , then either both series converge or both diverge.

Given series is $\displaystyle{\sum_{{{n}={1}}}^{\infty}}\frac{{5}}{{{{4}}^{{n}}+{1}}}$

Let the comparison series be $\displaystyle{\sum_{{{n}={1}}}^{\infty}}\frac{{1}}{{{4}}^{{n}}}={\sum_{{{n}={1}}}^{\infty}}{{\left(\frac{{1}}{{4}}\right)}}^{{n}}$

The comparison series $\displaystyle{\sum_{{{n}={1}}}^{\infty}}{{\left(\frac{{1}}{{4}}\right)}}^{{n}}$ is a geometric series with $\displaystyle{r}=\frac{{1}}{{4}}<{1}$

A geometric series with ratio r converges if $\displaystyle{0}<{\left|{r}\right|}<{1}$

So, the comparison series which is a geometric series converges.

Now let's use the limit...

Given series is $\displaystyle{\sum_{{{n}={1}}}^{\infty}}\frac{{5}}{{{{4}}^{{n}}+{1}}}$

Let the comparison series be $\displaystyle{\sum_{{{n}={1}}}^{\infty}}\frac{{1}}{{{4}}^{{n}}}={\sum_{{{n}={1}}}^{\infty}}{{\left(\frac{{1}}{{4}}\right)}}^{{n}}$

The comparison series $\displaystyle{\sum_{{{n}={1}}}^{\infty}}{{\left(\frac{{1}}{{4}}\right)}}^{{n}}$ is a geometric series with $\displaystyle{r}=\frac{{1}}{{4}}<{1}$

A geometric series with ratio r converges if $\displaystyle{0}<{\left|{r}\right|}<{1}$

So, the comparison series which is a geometric series converges.

Now let's use the limit comparison test with:

$\displaystyle{a}_{{n}}=\frac{{5}}{{{{4}}^{{n}}+{1}}}$ and $\displaystyle{b}_{{n}}=\frac{{1}}{{{4}}^{{n}}}$

$\displaystyle\frac{{a}_{{n}}}{{b}_{{n}}}=\frac{{\frac{{5}}{{{{4}}^{{n}}+{1}}}}}{{\frac{{1}}{{{4}}^{{n}}}}}$

$\displaystyle\frac{{a}_{{n}}}{{{b}}^{{n}}}=\frac{{{5}\cdot{{4}}^{{n}}}}{{{{4}}^{{n}}+{1}}}$

$\displaystyle\frac{{a}_{{n}}}{{b}_{{n}}}=\frac{{{5}\cdot{{4}}^{{n}}}}{{{{4}}^{{n}}{\left({1}+\frac{{1}}{{{4}}^{{n}}}\right)}}}$

$\displaystyle\frac{{a}_{{n}}}{{b}_{{n}}}=\frac{{5}}{{{1}+\frac{{1}}{{{4}}^{{n}}}}}$

$\displaystyle\lim_{{{n}\to\infty}}\frac{{a}_{{n}}}{{b}_{{n}}}=\lim_{{{n}\to\infty}}\frac{{5}}{{{1}+\frac{{1}}{{{4}}^{{n}}}}}$

$\displaystyle={5}>{0}$

Since the comparison series $\displaystyle{\sum_{{{n}={1}}}^{\infty}}{{\left(\frac{{1}}{{4}}\right)}}^{{n}}$ converges, so the series $\displaystyle{\sum_{{{n}={1}}}^{\infty}}\frac{{5}}{{{{4}}^{{n}}+{1}}}$ as well ,converges as per the limit comparison test.

New Notes Blog

Wednesday, 3 September 2014

$\displaystyle{\sum_{{{n}={1}}}^{\infty}}\frac{{5}}{{{{4}}^{{n}}+{1}}}$ Use the Limit Comparison Test to determine the convergence or divergence of the series.

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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, 3 September 2014

Use the Limit Comparison Test to determine the convergence or divergence of the series.

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?

$\displaystyle{\sum_{{{n}={1}}}^{\infty}}\frac{{5}}{{{{4}}^{{n}}+{1}}}$ Use the Limit Comparison Test to determine the convergence or divergence of the series.

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