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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

sum from n=1 to infinity of (1/(n+1))^2

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

converges

Solution steps

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

Simplify

(\frac{1}{n + 1})^{2} : \frac{1}{( n + 1 ) ^{2}}

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

Apply Series Comparison Test:

converges

= converges

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