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

Saturday, 25 October 2014

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

We may apply the Ratio Test to determine the convergence or divergence of the series $\displaystyle{\sum_{{{n}={0}}}^{\infty}}\frac{{{\left(-{1}\right)}}^{{n}}}{{{n}!}}$ .

In Ratio test, we determine the limit as:

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

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 Ratio Test to determine the convergence or divergence of the series $\displaystyle{\sum_{{{n}={0}}}^{\infty}}\frac{{{\left(-{1}\right)}}^{{n}}}{{{n}!}}$ .

In Ratio test, we determine the limit as:

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

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 series $\displaystyle{\sum_{{{n}={0}}}^{\infty}}\frac{{{\left(-{1}\right)}}^{{n}}}{{{n}!}}$ , we have $\displaystyle{a}_{{n}}=\frac{{{\left(-{1}\right)}}^{{n}}}{{{n}!}}$ .

Then, we may let $\displaystyle{a}_{{{n}+{1}}}=\frac{{{\left(-{1}\right)}}^{{{n}+{1}}}}{{{\left({n}+{1}\right)}!}}$

We set up the limit as:

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

To simplify the function, we flip the bottom and proceed to multiplication:

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

Apply Law of Exponent: $\displaystyle{{x}}^{{{n}+{m}}}={{x}}^{{n}}\cdot{{x}}^{{m}}$ and $\displaystyle{\left({n}+{1}\right)}!={n}!{\left({n}+{1}\right)}$

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

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

$\displaystyle{\left|\frac{{{\left(-{1}\right)}}^{{1}}}{{{n}+{1}}}\right|}$

$\displaystyle={\left|-\frac{{1}}{{{n}+{1}}}\right|}$

$\displaystyle=\frac{{1}}{{{n}+{1}}}$

Applying $\displaystyle{\left|\frac{{\frac{{{\left(-{1}\right)}}^{{{n}+{1}}}}{{{\left({n}+{1}\right)}!}}}}{{\frac{{{\left(-{1}\right)}}^{{n}}}{{{n}!}}}}\right|}=\frac{{1}}{{{n}+{1}}}$ , we get:

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

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

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

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

$\displaystyle={0}$

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

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

Notes

Saturday, 25 October 2014

$\displaystyle{\sum_{{{n}={0}}}^{\infty}}\frac{{{\left(-{1}\right)}}^{{n}}}{{{n}!}}$ 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?

Saturday, 25 October 2014

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?

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

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