New Notes Blog: `sum_(n=0)^oo 3/5^n` Determine the convergence or divergence of the series.

Tuesday, 26 August 2014

$\displaystyle{\sum_{{{n}={0}}}^{\infty}}\frac{{3}}{{{5}}^{{n}}}$ Determine the convergence or divergence of the series.

Recall that infinite series converges to single finite value $\displaystyle{S}$ if the limit if the partial sum $\displaystyle{S}_{{n}}$ as n approaches $\displaystyle\infty$ converges to $\displaystyle{S}$ . We follow it in a formula:

$\displaystyle\lim_{{{n}-\gt\infty}}{S}_{{n}}={\sum_{{{n}={1}}}^{\infty}}{a}_{{n}}={S}$ .

To evaluate the $\displaystyle{\sum_{{{n}={0}}}^{\infty}}\frac{{3}}{{{5}}^{{n}}}$ , we may express it in a form:

$\displaystyle{\sum_{{{n}={0}}}^{\infty}}\frac{{3}}{{{5}}^{{n}}}={\sum_{{{n}={0}}}^{\infty}}{3}\cdot{\left(\frac{{1}}{{{5}}^{{n}}}\right)}$

$\displaystyle={\sum_{{{n}={0}}}^{\infty}}{3}\cdot{{\left(\frac{{1}}{{5}}\right)}}^{{n}}$

This resembles form of geometric series with...

$\displaystyle\lim_{{{n}-\gt\infty}}{S}_{{n}}={\sum_{{{n}={1}}}^{\infty}}{a}_{{n}}={S}$ .

To evaluate the $\displaystyle{\sum_{{{n}={0}}}^{\infty}}\frac{{3}}{{{5}}^{{n}}}$ , we may express it in a form:

$\displaystyle{\sum_{{{n}={0}}}^{\infty}}\frac{{3}}{{{5}}^{{n}}}={\sum_{{{n}={0}}}^{\infty}}{3}\cdot{\left(\frac{{1}}{{{5}}^{{n}}}\right)}$

$\displaystyle={\sum_{{{n}={0}}}^{\infty}}{3}\cdot{{\left(\frac{{1}}{{5}}\right)}}^{{n}}$

This resembles form of geometric series with an index shift: $\displaystyle{\sum_{{{n}={0}}}^{\infty}}{a}\cdot{{r}}^{{n}}$ .

By comparing " $\displaystyle{3}\cdot{{\left(\frac{{1}}{{5}}\right)}}^{{n}}$ " with " $\displaystyle{a}\cdot{{r}}^{{n}}$ ", we determine the corresponding values: $\displaystyle{a}={3}$ and $\displaystyle{r}=\frac{{1}}{{5}}$ or $\displaystyle{0.2}$ .

The convergence test for the geometric series follows the conditions:

a) If $\displaystyle{\left|{r}\right|}\lt{1}$ or $\displaystyle-{1}\lt{r}\lt{1}$ then the geometric series converges to $\displaystyle{\sum_{{{n}={0}}}^{\infty}}{a}\cdot{{r}}^{{n}}=\frac{{a}}{{{1}-{r}}}$ .

b) If $\displaystyle{\left|{r}\right|}\geq{1}$ then the geometric series diverges.

The $\displaystyle{r}=\frac{{1}}{{5}}$ or $\displaystyle{0.2}$ from the given infinite series falls within the condition $\displaystyle{\left|{r}\right|}\lt{1}$ since $\displaystyle{\left|\frac{{1}}{{5}}\right|}\lt{1}$ or $\displaystyle{\left|{0.2}\right|}\lt{1}$ . Therefore, we may conclude that $\displaystyle{\sum_{{{n}={0}}}^{\infty}}\frac{{3}}{{{5}}^{{n}}}$ is a convergent series.

By applying the formula: $\displaystyle{\sum_{{{n}={0}}}^{\infty}}{a}\cdot{{r}}^{{n}}=\frac{{a}}{{{1}-{r}}}$ , we determine that the given geometric series will converge to a value:

$\displaystyle{\sum_{{{n}={0}}}^{\infty}}\frac{{3}}{{{5}}^{{n}}}={\sum_{{{n}={0}}}^{\infty}}{3}\cdot{{\left(\frac{{1}}{{5}}\right)}}^{{n}}$