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

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

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

converges

Solution steps

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

Apply p-Series Test:

converges

= converges

n = 0 \sum \infty \frac{3 n + 1}{2 n}

n = 1 \sum \infty \frac{1}{ln ( 4 ^{n} )}

n = 1 \sum \infty \frac{3}{n ^{5}}

n = 0 \sum \infty \frac{n ^{2}}{1 + 3 n}

n = 0 \sum \infty (\frac{11}{3})^{n}

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