What is the sum from n=2 to infinity of (e^n)/(pi^n) ?

The sum from n=2 to infinity of (e^n)/(pi^n) is converges

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

sum from n=2 to infinity of (e^n)/(pi^n)

n = 2 \sum \infty \frac{e ^{n}}{π ^{n}}

converges

Solution steps

n = 2 \sum \infty \frac{e ^{n}}{π ^{n}}

Apply Series Ratio Test:

converges

= converges

n = 0 \sum \infty 2^{- n}

n = 0 \sum \infty (0.3)^{n}

n = 0 \sum \infty (1 + \frac{2}{5 n})^{n}

n = 1 \sum \infty \frac{n}{2 n ^{2} - 3}

n = 0 \sum \infty (\frac{7}{9})^{n}

What is the sum from n=2 to infinity of (e^n)/(pi^n) ?
The sum from n=2 to infinity of (e^n)/(pi^n) is converges