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

Wednesday, 11 May 2016

$\displaystyle{\sum_{{{n}={1}}}^{\infty}}{{\left(\frac{{\ln{{n}}}}{{n}}\right)}}^{{n}}$ 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...

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 $\displaystyle{\sum_{{{n}={1}}}^{\infty}}{{\left(\frac{{\ln{{\left({n}\right)}}}}{{n}}\right)}}^{{n}}$ , we have $\displaystyle{a}_{{n}}={{\left(\frac{{\ln{{\left({n}\right)}}}}{{n}}\right)}}^{{n}}$ .

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

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

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

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

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

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

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

Evaluate the limit using direct substitution: $\displaystyle{n}=\infty$ .

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

When the limit value is indeterminate $\displaystyle{\left(\frac{\infty}{\infty}\right)}$ , we may apply L'Hospital's Rule:

$\displaystyle\lim_{{{x}-\gt{a}}}\frac{{{f{{\left({x}\right)}}}}}{{{g{{\left({x}\right)}}}}}=\lim_{{{x}-\gt{a}}}\frac{{{f{'}}{\left({x}\right)}}}{{{g{'}}{\left({x}\right)}}}$ .

Let: $\displaystyle{f{{\left({n}\right)}}}={\ln{{\left({n}\right)}}}$ then

$\displaystyle{g{{\left({n}\right)}}}={n}$ then $\displaystyle{g{'}}{\left({n}\right)}={1}$ .

Then, the limit becomes:

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