What is the sum from n=1 to infinity}(x^{n-1 of)/(sqrt(x)) ?

The sum from n=1 to infinity}(x^{n-1 of)/(sqrt(x)) is Interval of convergence is-1<x<1

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sum from n=1 to infinity}(x^{n-1 of)/(sqrt(x))

n = 1 \sum \infty \frac{x ^{n - 1}}{x}

Interval of convergence is - 1 < x < 1

Solution steps

n = 1 \sum \infty \frac{x ^{n - 1}}{x}

Use the Ratio Test to compute the convergence interval

Interval of convergence is - 1 < x < 1

x \to 1 lim (- 1 x^{- 1 - 1} sin (\frac{1}{x}) - x^{- 1 - 2} cos (\frac{1}{x}))

2 t + 3

y = x, y = x^{2}

\int \frac{( x ^{2} )}{2 x - x ^{2}} d x

f (x) = 4 x + \frac{6}{x x}

What is the sum from n=1 to infinity}(x^{n-1 of)/(sqrt(x)) ?
The sum from n=1 to infinity}(x^{n-1 of)/(sqrt(x)) is Interval of convergence is-1<x<1