What is the sum from n=1 to infinity of 3/n-3/(n+1) ?

The sum from n=1 to infinity of 3/n-3/(n+1) is 3

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 of 3/n-3/(n+1)

n = 1 \sum \infty \frac{3}{n} - \frac{3}{n + 1}

3

Solution steps

n = 1 \sum \infty \frac{3}{n} - \frac{3}{n + 1}

Apply Telescoping Series Test:

3

= 3

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

n = 1 \sum \infty \frac{3 ^{n}}{e ^{n}}

n = 0 \sum \infty 9 (\frac{8}{9})^{n}

n = 10 \sum \infty \frac{1}{n ln ( n )}

n = 1 \sum \infty 0. 1^{n}

What is the sum from n=1 to infinity of 3/n-3/(n+1) ?
The sum from n=1 to infinity of 3/n-3/(n+1) is 3