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

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

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

converges

Solution steps

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

Simplify

\frac{1}{n} - \frac{1}{n + 2} : \frac{2}{n ( n + 2 )}

= n = 3 \sum \infty \frac{2}{n ( n + 2 )}

Apply the constant multiplication rule:

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

= 2 \cdot n = 3 \sum \infty \frac{1}{n ( n + 2 )}

Apply Series Comparison Test:

converges

= 2 converges

= converges

n = 0 \sum \infty \frac{n}{6 n + 1}

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

n = 0 \sum \infty \frac{i ^{2 n}}{3 n}

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

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

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