Notes: `sum_(n=1)^oo (-1)^n/3^n` Determine the convergence or divergence of the series.

Monday, 28 April 2014

$\displaystyle{\sum_{{{n}={1}}}^{\infty}}\frac{{{\left(-{1}\right)}}^{{n}}}{{{3}}^{{n}}}$ Determine the convergence or divergence of the series.

To determine the convergence or divergence of the series $\displaystyle{\sum_{{{n}={1}}}^{\infty}}\frac{{{\left(-{1}\right)}}^{{n}}}{{{3}}^{{n}}}$ , we may apply the Ratio Test.

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 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 given series $\displaystyle{\sum_{{{n}={1}}}^{\infty}}\frac{{{\left(-{1}\right)}}^{{n}}}{{{3}}^{{n}}}$ ...

To determine the convergence or divergence of the series $\displaystyle{\sum_{{{n}={1}}}^{\infty}}\frac{{{\left(-{1}\right)}}^{{n}}}{{{3}}^{{n}}}$ , we may apply the Ratio Test.

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 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 given series $\displaystyle{\sum_{{{n}={1}}}^{\infty}}\frac{{{\left(-{1}\right)}}^{{n}}}{{{3}}^{{n}}}$ , we have $\displaystyle{a}_{{n}}=\frac{{{\left(-{1}\right)}}^{{n}}}{{{3}}^{{n}}}$ .

Then, $\displaystyle{a}_{{{n}+{1}}}=\frac{{{\left(-{1}\right)}}^{{{n}+{1}}}}{{{3}}^{{{n}+{1}}}}$ .

We set up the limit as:

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

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

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

Apply the Law of Exponent: $\displaystyle{{x}}^{{{n}+{m}}}={{x}}^{{n}}\cdot{{x}}^{{m}}$ . It becomes:

$\displaystyle{\left|\frac{{{{\left(-{1}\right)}}^{{n}}{{\left(-{1}\right)}}^{{1}}}}{{{{3}}^{{n}}\cdot{{3}}^{{1}}}}\cdot\frac{{{3}}^{{n}}}{{{\left(-{1}\right)}}^{{n}}}\right|}$

Cancel out common factors $\displaystyle{{\left(-{1}\right)}}^{{n}}$ and $\displaystyle{\left({{3}}^{{n}}\right)}$ .

$\displaystyle{\left|\frac{{{\left(-{1}\right)}}^{{1}}}{{{3}}^{{1}}}\right|}$

Simplify:

$\displaystyle{\left|\frac{{{\left(-{1}\right)}}^{{1}}}{{{3}}^{{1}}}\right|}={\left|\frac{{-{1}}}{{3}}\right|}$

$\displaystyle={\left|-\frac{{1}}{{3}}\right|}$

$\displaystyle=\frac{{1}}{{3}}$

Applying $\displaystyle{\left|\frac{{\frac{{{\left(-{1}\right)}}^{{{n}+{1}}}}{{{3}}^{{{n}+{1}}}}}}{{\frac{{{\left(-{1}\right)}}^{{n}}}{{{3}}^{{n}}}}}\right|}=\frac{{1}}{{3}}$ , we get:

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