Notes: `sum_(n=2)^oo n/(nlnn)^n` Use the Root Test to determine the convergence or divergence of the series.

Thursday, 28 August 2014

$\displaystyle{\sum_{{{n}={2}}}^{\infty}}\frac{{n}}{{{\left({n}{\ln{{n}}}\right)}}^{{n}}}$ 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.

For the given series $\displaystyle{\sum_{{{n}={2}}}^{\infty}}\ldots$

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.

For the given series $\displaystyle{\sum_{{{n}={2}}}^{\infty}}\frac{{n}}{{{\left({n}{\ln{{\left({n}\right)}}}\right)}}^{{n}}}$ , we have $\displaystyle{a}_{{n}}=\frac{{n}}{{{\left({n}{\ln{{\left({n}\right)}}}\right)}}^{{n}}}$ .

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

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

Apply Law of Exponent: $\displaystyle{{\left(\frac{{x}}{{y}}\right)}}^{{n}}=\frac{{{x}}^{{n}}}{{{y}}^{{n}}}$ and $\displaystyle{{\left({{x}}^{{n}}\right)}}^{{m}}={{x}}^{{{n}\cdot{m}}}$ .

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

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

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

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

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

Apply the limit property: $\displaystyle\lim_{{{x}-\gt{a}}}{\left[\frac{{{f{{\left({x}\right)}}}}}{{{g{{\left({x}\right)}}}}}\right]}=\frac{{\lim_{{{x}-\gt{a}}}{f{{\left({x}\right)}}}}}{{\lim_{{{x}-\gt{a}}}{g{{\left({x}\right)}}}}}.$

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

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

$\displaystyle={0}$

Note: $\displaystyle\lim_{{{n}-\gt\infty}}{{n}}^{{\frac{{1}}{{n}}}}={1}$ and

$\displaystyle\lim_{{{n}-\gt\infty}}{n}{\ln{{\left({n}\right)}}}=\infty{\ln{{\left(\infty\right)}}}$

$\displaystyle=\infty\cdot\infty$

$\displaystyle=\infty$

The limit value $\displaystyle{L}={0}$ satisfies the condition: $\displaystyle{L}\lt{1}$ since $\displaystyle{0}\lt{1}$ .

Conclusion: The series $\displaystyle{\sum_{{{n}={2}}}^{\infty}}\frac{{n}}{{{\left({n}{\ln{{\left({n}\right)}}}\right)}}^{{n}}}$ is absolutely convergent.

Notes

Thursday, 28 August 2014

$\displaystyle{\sum_{{{n}={2}}}^{\infty}}\frac{{n}}{{{\left({n}{\ln{{n}}}\right)}}^{{n}}}$ Use the Root Test to determine the convergence or divergence of the series.

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How are race, gender, and class addressed in Oliver Optic's Rich and Humble?

Thursday, 28 August 2014

Use the Root Test to determine the convergence or divergence of the series.

No comments:

Post a Comment

How are race, gender, and class addressed in Oliver Optic&#39;s Rich and Humble?

$\displaystyle{\sum_{{{n}={2}}}^{\infty}}\frac{{n}}{{{\left({n}{\ln{{n}}}\right)}}^{{n}}}$ Use the Root Test to determine the convergence or divergence of the series.

How are race, gender, and class addressed in Oliver Optic's Rich and Humble?