What is the sum from n=1 to infinity of 3/((4n+5)^3) ?

The sum from n=1 to infinity of 3/((4n+5)^3) 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 3/((4n+5)^3)

n = 1 \sum \infty \frac{3}{( 4 n + 5 ) ^{3}}

converges

Solution steps

n = 1 \sum \infty \frac{3}{( 4 n + 5 ) ^{3}}

Apply the constant multiplication rule:

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

= 3 \cdot n = 1 \sum \infty \frac{1}{( 4 n + 5 ) ^{3}}

Apply Series Comparison Test:

converges

= 3 converges

= converges

n = 1 \sum \infty 4 sin (\frac{2}{n})

r = 1 \sum \infty r (2 r - 1)

k = 0 \sum \infty (- \frac{1}{6})^{k}

k = 1 \sum \infty \frac{k}{e ^{3 k}}

n = 1 \sum \infty \frac{1}{6 ^{n} + n}

What is the sum from n=1 to infinity of 3/((4n+5)^3) ?
The sum from n=1 to infinity of 3/((4n+5)^3) is converges