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

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

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

converges

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Series Comparison Test:

converges

= 3 converges

= converges

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

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

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

n = 1 \sum \infty (\frac{99}{100})^{n}

n = 1 \sum \infty \frac{7}{n ^{3}}

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