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

Friday, 10 July 2015

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

To determine the convergence or divergence of a series $\displaystyle\sum{a}_{{n}}$ using Root test, we evaluate a limit as:

$\displaystyle\lim_{{{n}-\gt\infty}}{\sqrt[{{n}}]{{{\left|{a}_{{n}}\right|}}}}={L}$

$\displaystyle\lim_{{{n}-\gt\infty}}{{\left|{a}_{{n}}\right|}}^{{\frac{{1}}{{n}}}}={L}$

Then, we follow the conditions:

a) $\displaystyle{L}\lt{1}$ then the series is absolutely convergent.

b) $\displaystyle{L}\gt{1}$ then the series is divergent.

c) $\displaystyle{L}={1}$ or does not exist then the test is inconclusive. The series may be divergent, conditionally convergent, or absolutely convergent.

We may...

To determine the convergence or divergence of a series $\displaystyle\sum{a}_{{n}}$ using Root test, we evaluate a limit as:

$\displaystyle\lim_{{{n}-\gt\infty}}{\sqrt[{{n}}]{{{\left|{a}_{{n}}\right|}}}}={L}$

$\displaystyle\lim_{{{n}-\gt\infty}}{{\left|{a}_{{n}}\right|}}^{{\frac{{1}}{{n}}}}={L}$

Then, we follow the conditions:

a) $\displaystyle{L}\lt{1}$ then the series is absolutely convergent.

b) $\displaystyle{L}\gt{1}$ then the series is divergent.

c) $\displaystyle{L}={1}$ or does not exist then the test is inconclusive. The series may be divergent, conditionally convergent, or absolutely convergent.

We may apply the Root Test to determine the convergence or divergence of the series $\displaystyle{\sum_{{{n}={1}}}^{\infty}}\frac{{1}}{{{5}}^{{n}}}.$

For the given series $\displaystyle{\sum_{{{n}={1}}}^{\infty}}\frac{{1}}{{{5}}^{{n}}},$ we have $\displaystyle{a}_{{n}}=\frac{{1}}{{{5}}^{{n}}}.$

Applying the Root test, we set-up the limit as:

$\displaystyle\lim_{{{n}-\gt\infty}}{{\left|\frac{{1}}{{{5}}^{{n}}}\right|}}^{{\frac{{1}}{{n}}}}=\lim_{{{n}-\gt\infty}}{{\left(\frac{{1}}{{{5}}^{{n}}}\right)}}^{{\frac{{1}}{{n}}}}$

Apply the Law of Exponents: $\displaystyle{{\left(\frac{{x}}{{y}}\right)}}^{{n}}={\left(\frac{{{x}}^{{n}}}{{{y}}^{{n}}}\right)}$ .

$\displaystyle\lim_{{{n}-\gt\infty}}{{\left(\frac{{1}}{{{5}}^{{n}}}\right)}}^{{\frac{{1}}{{n}}}}=\lim_{{{n}-\gt\infty}}\frac{{{1}}^{{\frac{{1}}{{n}}}}}{{{5}}^{{{n}\cdot{\left(\frac{{1}}{{n}}\right)}}}}$

$\displaystyle=\lim_{{{n}-\gt\infty}}\frac{{{1}}^{{\frac{{1}}{{n}}}}}{{{5}}^{{\frac{{n}}{{n}}}}}$

$\displaystyle=\lim_{{{n}-\gt\infty}}\frac{{{1}}^{{\frac{{1}}{{n}}}}}{{{5}}^{{1}}}$

$\displaystyle=\lim_{{{n}-\gt\infty}}\frac{{{1}}^{{\frac{{1}}{{n}}}}}{{5}}$

Evaluate the limit.

$\displaystyle\lim_{{{n}-\gt\infty}}\frac{{{1}}^{{\frac{{1}}{{n}}}}}{{5}}=\frac{{1}}{{5}}\lim_{{{n}-\gt\infty}}{{1}}^{{\frac{{1}}{{n}}}}$

$\displaystyle=\frac{{1}}{{5}}\cdot{{1}}^{{\frac{{1}}{\infty}}}$

$\displaystyle=\frac{{1}}{{5}}\cdot{{1}}^{{0}}$

$\displaystyle=\frac{{1}}{{5}}\cdot{1}$

$\displaystyle=\frac{{1}}{{5}}$

The limit value $\displaystyle{L}=\frac{{1}}{{5}}$ satisfies the condition: $\displaystyle{L}\lt{1}$ .

Thus, the series $\displaystyle{\sum_{{{n}={1}}}^{\infty}}\frac{{1}}{{{5}}^{{n}}}$ is absolutely convergent.

New Notes Blog

Friday, 10 July 2015

$\displaystyle{\sum_{{{n}={1}}}^{\infty}}\frac{{1}}{{{5}}^{{n}}}$ Use the Root 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?

Friday, 10 July 2015

Use the Root 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?

$\displaystyle{\sum_{{{n}={1}}}^{\infty}}\frac{{1}}{{{5}}^{{n}}}$ Use the Root 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?