What is the sum from n=7 to infinity of 1/((2n-1)^2) ?

The sum from n=7 to infinity of 1/((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=7 to infinity of 1/((2n-1)^2)

n = 7 \sum \infty \frac{1}{( 2 n - 1 ) ^{2}}

converges

Solution steps

n = 7 \sum \infty \frac{1}{( 2 n - 1 ) ^{2}}

Apply Series Comparison Test:

converges

= converges

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

n = 1 \sum \infty (- \frac{4}{5})^{n - 1}

n = 0 \sum \infty (\frac{3}{4})^{- n}

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

n = 1 \sum \infty \frac{1}{4 + 3 ^{n}}

What is the sum from n=7 to infinity of 1/((2n-1)^2) ?
The sum from n=7 to infinity of 1/((2n-1)^2) is converges