Question

Find all values of x for which the series converges. (Enter your answer using interval notation.) [infinity] 9 x − 7 9 n n = 0 For these values of x, write the sum of the series as a function of x.

Answer 1

Answer:

The series converges to   $\frac{1}{1-9x}$     for   $\frac{-1}{9} < x < \frac{1}{9}$                

Step-by-step explanation:

Given the series is     $\sum_{n=0}^{\infty} 9^n x^n$

We have to find the values of x for which the series converges.

We know,

$\sum_{n=0}^{\infty} ar^{n-1}$ converges to  (a) / (1-r) if r < 1

Otherwise the series will diverge.

Here, $\sum_{n=0}^{\infty} 9^n x^n = \sum_{n=0}^{\infty} (9x)^{n}$ is a geometric series with |r| = | 9x |

And it converges for |9x| < 1

Hence, the given series gets converge for $\frac{-1}{9} < x < \frac{1}{9}$

And geometric series converges to $\frac{a}{1-r}$

Here, a = 1 and r = 9x

Therefore, $\frac{a}{1-r} = \frac{1}{1-9x}$

Hence, the given series converges to   $\frac{1}{1-9x}$     for   $\frac{-1}{9} < x < \frac{1}{9}$