Question

# Find the sum of the convergent series. sum_{n=0}^infty5(frac23)^n

Series
Find the sum of the convergent series.
$$\sum_{n=0}^\infty5(\frac23)^n$$

2021-02-16
To find the sum of the convergent series: $$\sum_{n=0}^\infty5(\frac23)^n$$
Solution:
Expanding the given series, we get
$$\sum_{n=0}^\infty5(\frac23)^n=5(\frac23)^1+5(\frac23)^2+5(\frac23)^3+...$$
$$=5[(\frac23)^1+(\frac23)^2+(\frac23)^3+...]$$
Now, taking the series $$(\frac23)^1+(\frac23)^2+(\frac23)^3+...$$
Here,we can find that sequence is in geometric progression.
First term is $$a_1=\frac23$$
Common ratio is:
$$r=\frac{(\frac23)^2}{(\frac23)}$$
$$=\frac23$$
Sum of infinite terms of G.P. is given as:
$$S=\frac{a_1}{1-r}$$
Sum of the sequence $$(\frac23)^1+(\frac23)^2+(\frac23)^3+...$$ will be:
$$S=\frac{\frac23}{1-\frac23}$$
$$=\frac{\frac23}{\frac13}$$
$$=2$$
Now, sum of the series $$\sum_{n=0}^\infty5(\frac23)^n$$ will be:
$$\sum_{n=0}^\infty5(\frac23)^n=5[(\frac23)^1+(\frac23)^2+(\frac23)^3+...]$$
$$=5\cdot2$$
$$=10$$
Hence, required sum is 10.