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Sagot :
The series converges to 1/(1-9x) for -1/9<x<1/9
Given the series is ∑ [tex]9x^{n} x^{n}[/tex]
We have to find the values of x for which the series converges.
We know,
∑ [tex]ar^{n-1}[/tex] converges to (a) / (1-r) if r < 1
Otherwise the series will diverge.
Here, ∑ [tex](9x)^{n}[/tex] 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
https://brainly.com/question/15415793
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