 # Find the sum of the convergent series. sum_{n=0}^infty5(frac23)^n Khadija Wells 2021-02-15 Answered
Find the sum of the convergent series.
$\sum _{n=0}^{\mathrm{\infty }}5\left(\frac{2}{3}{\right)}^{n}$
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it toroztatG
To find the sum of the convergent series: $\sum _{n=0}^{\mathrm{\infty }}5\left(\frac{2}{3}{\right)}^{n}$
Solution:
Expanding the given series, we get
$\sum _{n=0}^{\mathrm{\infty }}5\left(\frac{2}{3}{\right)}^{n}=5\left(\frac{2}{3}{\right)}^{1}+5\left(\frac{2}{3}{\right)}^{2}+5\left(\frac{2}{3}{\right)}^{3}+...$
$=5\left[\left(\frac{2}{3}{\right)}^{1}+\left(\frac{2}{3}{\right)}^{2}+\left(\frac{2}{3}{\right)}^{3}+...\right]$
Now, taking the series $\left(\frac{2}{3}{\right)}^{1}+\left(\frac{2}{3}{\right)}^{2}+\left(\frac{2}{3}{\right)}^{3}+...$
Here,we can find that sequence is in geometric progression.
First term is ${a}_{1}=\frac{2}{3}$
Common ratio is:
$r=\frac{\left(\frac{2}{3}{\right)}^{2}}{\left(\frac{2}{3}\right)}$
$=\frac{2}{3}$
Sum of infinite terms of G.P. is given as:
$S=\frac{{a}_{1}}{1-r}$
Sum of the sequence $\left(\frac{2}{3}{\right)}^{1}+\left(\frac{2}{3}{\right)}^{2}+\left(\frac{2}{3}{\right)}^{3}+...$ will be:
$S=\frac{\frac{2}{3}}{1-\frac{2}{3}}$
$=\frac{\frac{2}{3}}{\frac{1}{3}}$
$=2$
Now, sum of the series $\sum _{n=0}^{\mathrm{\infty }}5\left(\frac{2}{3}{\right)}^{n}$ will be:
$\sum _{n=0}^{\mathrm{\infty }}5\left(\frac{2}{3}{\right)}^{n}=5\left[\left(\frac{2}{3}{\right)}^{1}+\left(\frac{2}{3}{\right)}^{2}+\left(\frac{2}{3}{\right)}^{3}+...\right]$
$=5\cdot 2$
$=10$
Hence, required sum is 10.
###### Not exactly what you’re looking for? Jeffrey Jordon