What is the sum from n=1 to infinity}(e^{2/n of)/n ?

The sum from n=1 to infinity}(e^{2/n of)/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=1 to infinity}(e^{2/n of)/n

n = 1 \sum \infty \frac{e ^{\frac{2}{n}}}{n}

converges

Solution steps

n = 1 \sum \infty \frac{e ^{\frac{2}{n}}}{n}

Apply Series Integral Test:

converges

= converges

n = 1 \sum \infty \frac{1}{n ^{9}}

n = 1 \sum \infty \frac{9}{n} - \frac{9}{n + 1}

n = 1 \sum \infty \frac{6}{3 n ^{2} + 2}

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

n = 1 \sum \infty 1 - \frac{1}{n}

What is the sum from n=1 to infinity}(e^{2/n of)/n ?
The sum from n=1 to infinity}(e^{2/n of)/n is converges