What is the sum from n=0 to infinity of (-1)^nx(x/4)^{2n} ?

The sum from n=0 to infinity of (-1)^nx(x/4)^{2n} is Interval of convergence is-4<x<4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

sum from n=0 to infinity of (-1)^nx(x/4)^{2n}

n = 0 \sum \infty (- 1)^{n} x (\frac{x}{4})^{2 n}

Interval of convergence is - 4 < x < 4

Solution steps

n = 0 \sum \infty (- 1)^{n} x (\frac{x}{4})^{2 n}

Use the Ratio Test to compute the convergence interval

Interval of convergence is - 4 < x < 4

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

y = sin (sin (sin (x)))

x \to 5 lim (x 29 - x^{2})

e^{t} sin (3 t)

\int sin^{6} (2 x) cos^{3} (2 x) d x

What is the sum from n=0 to infinity of (-1)^nx(x/4)^{2n} ?
The sum from n=0 to infinity of (-1)^nx(x/4)^{2n} is Interval of convergence is-4<x<4