What is the sum from n=1 to infinity of 1/(4n^2-49) ?

The sum from n=1 to infinity of 1/(4n^2-49) 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/(4n^2-49)

n = 1 \sum \infty \frac{1}{4 n ^{2} - 49}

converges

Solution steps

n = 1 \sum \infty \frac{1}{4 n ^{2} - 49}

Apply Series Integral Test:

converges

= converges

n = 1 \sum \infty 6 n^{2} e^{- n^{3}}

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

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

n = 1 \sum \infty n e^{- 5 n^{2}}

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

What is the sum from n=1 to infinity of 1/(4n^2-49) ?
The sum from n=1 to infinity of 1/(4n^2-49) is converges