Question

Find all values of x for which the series converges. (enter your answer using interval notation. ) [infinity] (4x)n n = 1

Answer 1

The series converges to  1/(1-9x) for -1/9<x<1/9                  

Given the series is  ∑ $9x^{n} x^{n}$

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

We know,                

∑ $ar^{n-1}$    converges to  (a) / (1-r) if r < 1

Otherwise the series will diverge.

Here, ∑ $(9x)^{n}$ is a geometric series with |r| = | 9x |

And it converges for |9x| < 1

Hence, the given series gets converge for -1/9<x<1/9

And geometric series converges to a/(1-r)

Here, a = 1 and r = 9x

Therefore, a/(1-r) = 1/(1-9x)

Hence, the given series converges to   1/1-9x  for  -1/9<x<1/9

For more information about convergence of series, visit

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