What is the sum from n=0 to infinity of (1^n)/(2n+1) ?

The sum from n=0 to infinity of (1^n)/(2n+1) is diverges

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^n)/(2n+1)

n = 0 \sum \infty \frac{1 ^{n}}{2 n + 1}

diverges

Solution steps

n = 0 \sum \infty \frac{1 ^{n}}{2 n + 1}

Simplify

= n = 0 \sum \infty \frac{1}{2 n + 1}

Apply Series Integral Test:

diverges

= diverges

\frac{d}{d x} (x^{8} sin (x))

\frac{\partial}{\partial x} (\frac{x ^{4}}{y})

\frac{\partial}{\partial y} (6 + x ln (x y - 5))

\int 3 sin^{3} (x) cos^{7} (x) d x

x = 16 - y^{2}, x = y^{2} - 16

What is the sum from n=0 to infinity of (1^n)/(2n+1) ?
The sum from n=0 to infinity of (1^n)/(2n+1) is diverges