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

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

n = 1 \sum \infty \frac{1}{n} - \frac{1}{n + 5}

converges

Solution steps

n = 1 \sum \infty \frac{1}{n} - \frac{1}{n + 5}

Simplify

\frac{1}{n} - \frac{1}{n + 5} : \frac{5}{n ( n + 5 )}

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

Apply the constant multiplication rule:

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

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

Apply Series Comparison Test:

converges

= 5 converges

= converges

n = 1 \sum \infty (\frac{8}{5})^{n}

n = 0 \sum \infty 20 (\frac{1}{4})^{n}

k = 2 \sum \infty (\frac{1}{3})^{2 k}

n = 0 \sum \infty (\frac{- 3}{4})^{n}

n = 1 \sum \infty \frac{5}{n ^{2} + 1}

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