What is the sum from n=1 to infinity of 5/(3n^3+1) ?

The sum from n=1 to infinity of 5/(3n^3+1) 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 5/(3n^3+1)

n = 1 \sum \infty \frac{5}{3 n ^{3} + 1}

converges

Solution steps

n = 1 \sum \infty \frac{5}{3 n ^{3} + 1}

Apply the constant multiplication rule:

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

= 5 \cdot n = 1 \sum \infty \frac{1}{3 n ^{3} + 1}

Apply Series Comparison Test:

converges

= 5 converges

= converges

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

n = 1 \sum \infty 5 e^{- 0.5 n}

n = 0 \sum \infty \frac{7}{n + 3}

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

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

What is the sum from n=1 to infinity of 5/(3n^3+1) ?
The sum from n=1 to infinity of 5/(3n^3+1) is converges