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

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

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

n = 1 \sum \infty (- x)^{n + 1}

Interval of convergence is - 1 < x < 1

Solution steps

n = 1 \sum \infty (- x)^{n + 1}

Use the Ratio Test to compute the convergence interval

Interval of convergence is - 1 < x < 1

9 ln (5 - 8 x^{2})

\frac{d y}{d t} = 3 + 2 y

y = x^{3} - 3 x - 4

\frac{d y}{d x} = \frac{2 x ( y - 1 )}{x ^{2} + 3}, y (1) = 9

\int \frac{x}{( x ^{2} + 1 ) ^{4}} d x

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