What is the sum from n=2 to infinity of x^n ?

The sum from n=2 to infinity of x^n is Radius of convergence is 1, Interval of convergence is-1<x<1

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sum from n=2 to infinity of x^n

n = 2 \sum \infty x^{n}

Radius of convergence is 1, Interval of convergence is - 1 < x < 1

Solution steps

n = 2 \sum \infty x^{n}

Use the Root Test to compute the convergence interval

Radius of convergence is 1, Interval of convergence is - 1 < x < 1

\frac{d y}{d x} = - \frac{sin ( x )}{y}

\int cos (r^{2}) d r

x y^{^{'}} = (1 - x) y + x^{2} + x^{3}

y^{^{''}} + 4 y^{^{'}} = cos^{2} (x)

\int - \frac{1}{10 - x} d x

What is the sum from n=2 to infinity of x^n ?
The sum from n=2 to infinity of x^n is Radius of convergence is 1, Interval of convergence is-1<x<1