What is the sum from n=1 to infinity of 4/(n^2) ?

The sum from n=1 to infinity of 4/(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 4/(n^2)

n = 1 \sum \infty \frac{4}{n ^{2}}

converges

Solution steps

n = 1 \sum \infty \frac{4}{n ^{2}}

Apply the constant multiplication rule:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

= 4 \cdot n = 1 \sum \infty \frac{1}{n ^{2}}

Apply p-Series Test:

converges

= 4 converges

= converges

n = 1 \sum \infty 2 (\frac{1}{3})^{n - 1}

n = 1 \sum \infty \frac{9}{n ^{2} + 25}

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

n = 0 \sum \infty (- \frac{3}{7})^{n}

n = 1 \sum \infty 9 (0.8)^{n - 1}

What is the sum from n=1 to infinity of 4/(n^2) ?
The sum from n=1 to infinity of 4/(n^2) is converges