Question

#3 n-1 Determine whether the series 2 35 n=1 converges or diverges. If it converges, find its sum. #41 00 n-1 Analyze for convergence (6) NO -0") . Find the sum in case the series converges. n=1

Answer 1

(3)

we are given

27 71-1 T1=1

we can expand it

$\sum_{n=1}^{\infty}\frac{2}{3}(\frac{7}{5})^{n-1}=\frac{2}{3}(\frac{7}{5})^{1-1}+\frac{2}{3}(\frac{7}{5})^{2-1}+\frac{2}{3}(\frac{7}{5})^{3-1}+\frac{2}{3}(\frac{7}{5})^{4-1}+........$

First term is

2 a 3

Common ratio is

$r=\frac{7}{5}$

Since,

so, this series is divergent......Answer

(4)

we are given series as

$\sum_{n=1}^{\infty}((\frac{2}{5})^{n-1}-(\frac{1}{4})^{n-1})$

we can expand it

$\sum_{n=1}^{\infty}((\frac{2}{5})^{n-1}-(\frac{1}{4})^{n-1})=\sum_{n=1}^{\infty}(\frac{2}{5})^{n-1}-\sum_{n=1}^{\infty}(\frac{1}{4})^{n-1}$

$\sum_{n=1}^{\infty}(\frac{2}{5})^{n-1}$

First term is

$a=1$

Common ratio is

$r=\frac{2}{5}$

now, we can find sum

$sum=\frac{a}{1-r}$

$sum=\frac{1}{1-\frac{2}{5}}$

$sum=\frac{5}{3}$

$\sum_{n=1}^{\infty}(\frac{1}{4})^{n-1}$

First term is

$a=1$

Common ratio is

$r=\frac{1}{4}$

now, we can find sum

$sum=\frac{a}{1-r}$

$sum=\frac{1}{1-\frac{1}{4}}$

$sum=\frac{4}{3}$

now, we can find sum

$\sum_{n=1}^{\infty}((\frac{2}{5})^{n-1}-(\frac{1}{4})^{n-1})=\frac{5}{3}-\frac{4}{3}$

$\sum_{n=1}^{\infty}((\frac{2}{5})^{n-1}-(\frac{1}{4})^{n-1})=\frac{1}{3}$..............Answer