What is the sum from n=0 to infinity of (e/(4pi))^n ?

The sum from n=0 to infinity of (e/(4pi))^n is (4pi)/(4pi-e)

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 (e/(4pi))^n

n = 0 \sum \infty (\frac{e}{4 π})^{n}

\frac{4 π}{4 π - e}

Solution steps

n = 0 \sum \infty (\frac{e}{4 π})^{n}

Apply Series Geometric Test:

\frac{4 π}{4 π - e}

= \frac{4 π}{4 π - e}

\int \frac{x ^{2} + 5 x + 4}{x ^{2} - 2 x + 1} d x

\frac{d}{d x} (tan (sec (cos (x))))

y^{4} - 4 y^{2} = x^{4} - 9 x^{2}, (- 3, 2)

(\frac{1}{y ^{2}} - \frac{3}{y ^{4}}) (y + 5 y^{3})

\frac{d y}{d x} = (e^{x - y})

What is the sum from n=0 to infinity of (e/(4pi))^n ?
The sum from n=0 to infinity of (e/(4pi))^n is (4pi)/(4pi-e)