What is the sum from n=0 to infinity of (1/n)^2 ?

The sum from n=0 to infinity of (1/n)^2 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=0 to infinity of (1/n)^2

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

converges

Solution steps

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

Simplify

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

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

Apply p-Series Test:

converges

= converges

k = 0 \sum \infty \frac{2 ^{k}}{k !}

n = 1 \sum \infty \frac{0. 5 ^{n}}{n}

n = 0 \sum \infty \frac{2 ^{2 n}}{n !}

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

n = 1 \sum \infty n! e^{- 8 n}

What is the sum from n=0 to infinity of (1/n)^2 ?
The sum from n=0 to infinity of (1/n)^2 is converges