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

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

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

converges

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Series Comparison Test:

converges

= 2 converges

= converges

x = 0 \sum \infty x

n = 0 \sum \infty 3 n - 3

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

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

n = 1 \sum \infty \frac{7}{6 n}

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