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WARRIOR [948]
3 years ago
5

Find all values of x for which the series converges. (Enter your answer using interval notation.) [infinity] 9 x − 7 9 n n = 0 F

or these values of x, write the sum of the series as a function of x.
Mathematics
1 answer:
makkiz [27]3 years ago
4 0

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} $

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