Notes: `sum_(n=0)^oo (-1)^n/((2n+1)!)` Determine the convergence or divergence of the series.

Friday, 21 October 2016

$\displaystyle{\sum_{{{n}={0}}}^{\infty}}\frac{{{\left(-{1}\right)}}^{{n}}}{{{\left({2}{n}+{1}\right)}!}}$ Determine the convergence or divergence of the series.

To determine the convergence or divergence of the series $\displaystyle{\sum_{{{n}={0}}}^{\infty}}\frac{{{\left(-{1}\right)}}^{{n}}}{{{\left({2}{n}+{1}\right)}!}}$ , we may apply ratio test.

In Ratio test, we determine the limit as:

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

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

Then ,we follow the conditions:

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

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

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

For...

To determine the convergence or divergence of the series $\displaystyle{\sum_{{{n}={0}}}^{\infty}}\frac{{{\left(-{1}\right)}}^{{n}}}{{{\left({2}{n}+{1}\right)}!}}$ , we may apply ratio test.

In Ratio test, we determine the limit as:

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

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

Then ,we follow the conditions:

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

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

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 series $\displaystyle{\sum_{{{n}={0}}}^{\infty}}\frac{{{\left(-{1}\right)}}^{{n}}}{{{\left({2}{n}+{1}\right)}!}}$ , we have:

$\displaystyle{a}_{{n}}=\frac{{{\left(-{1}\right)}}^{{n}}}{{{\left({2}{n}+{1}\right)}!}}$

Then,

$\displaystyle\frac{{1}}{{a}_{{n}}}=\frac{{{\left({2}{n}+{1}\right)}!}}{{{\left(-{1}\right)}}^{{n}}}$

$\displaystyle{a}_{{{n}+{1}}}=\frac{{{\left(-{1}\right)}}^{{{n}+{1}}}}{{{\left({2}{\left({n}+{1}\right)}+{1}\right)}!}}$

$\displaystyle=\frac{{{\left(-{1}\right)}}^{{{n}+{1}}}}{{{\left({2}{n}+{2}+{1}\right)}!}}$

$\displaystyle=\frac{{{\left(-{1}\right)}}^{{{n}+{1}}}}{{{\left({2}{n}+{3}\right)}!}}$

$\displaystyle=\frac{{{{\left(-{1}\right)}}^{{n}}\cdot{\left(-{1}\right)}}}{{{\left({2}{n}+{3}\right)}{\left({2}{n}+{2}\right)}{\left({\left({2}{n}+{1}\right)}!\right)}}}$

Applying the Ratio test on the power series, we set-up the limit as:

$\displaystyle\lim_{{{n}-\gt\infty}}{\left|\frac{{{{\left(-{1}\right)}}^{{n}}\cdot{\left(-{1}\right)}}}{{{\left({2}{n}+{3}\right)}{\left({2}{n}+{2}\right)}{\left({\left({2}{n}+{1}\right)}!\right)}}}\cdot\frac{{{\left({2}{n}+{1}\right)}!}}{{{\left(-{1}\right)}}^{{n}}}\right|}$

Cancel out common factors: $\displaystyle{{\left(-{1}\right)}}^{{n}}$ and $\displaystyle{\left({2}{n}+{1}\right)}!$ .

$\displaystyle\lim_{{{n}-\gt\infty}}{\left|\frac{{-{1}}}{{{\left({2}{n}+{3}\right)}{\left({2}{n}+{2}\right)}}}\right|}$

Evaluate the limit.

$\displaystyle\lim_{{{n}-\gt\infty}}{\left|\frac{{-{1}}}{{{\left({2}{n}+{3}\right)}{\left({2}{n}+{2}\right)}}}\right|}={\left|-{1}\right|}\lim_{{{n}-\gt\infty}}{\left|\frac{{1}}{{{\left({2}{n}+{3}\right)}{\left({2}{n}+{2}\right)}}}\right|}$

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

$\displaystyle={1}\cdot{0}$

$\displaystyle={0}$

The $\displaystyle{L}={0}$ satisfies ratio test condition: $\displaystyle{L}\lt{1}$ since $\displaystyle{0}\lt{1}$ .

Thus, the series $\displaystyle{\sum_{{{n}={0}}}^{\infty}}\frac{{{\left(-{1}\right)}}^{{n}}}{{{\left({2}{n}+{1}\right)}!}}$ is absolutely convergent.