What is the sum from n=1 to infinity of 1/(3n^2+8) ?

The sum from n=1 to infinity of 1/(3n^2+8) 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/(3n^2+8)

n = 1 \sum \infty \frac{1}{3 n ^{2} + 8}

converges

Solution steps

n = 1 \sum \infty \frac{1}{3 n ^{2} + 8}

Apply Series Comparison Test:

converges

= converges

k = 6 \sum \infty \frac{1}{k ^{2} - 5 k + 6}

n = 2 \sum \infty \frac{1}{4 n ^{3}}

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

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

n = 1 \sum \infty 0. 5^{n - 1}

What is the sum from n=1 to infinity of 1/(3n^2+8) ?
The sum from n=1 to infinity of 1/(3n^2+8) is converges