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

The sum from k=3 to infinity of 5/k-5/(k+1) is 5/3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

sum from k=3 to infinity of 5/k-5/(k+1)

k = 3 \sum \infty \frac{5}{k} - \frac{5}{k + 1}

\frac{5}{3}

Solution steps

k = 3 \sum \infty \frac{5}{k} - \frac{5}{k + 1}

Expand

\frac{5}{k} - \frac{5}{k + 1} : \frac{5}{k ^{2} + k}

= k = 3 \sum \infty \frac{5}{k ^{2} + k}

Apply the constant multiplication rule:

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

= 5 \cdot k = 3 \sum \infty \frac{1}{k ^{2} + k}

Apply Telescoping Series Test:

\frac{1}{3}

= 5 \cdot \frac{1}{3}

Simplify

5 \cdot \frac{1}{3} : \frac{5}{3}

= \frac{5}{3}

k = 0 \sum \infty (0.9)^{k}

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

n = 0 \sum \infty 0.99 9^{n}

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

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

What is the sum from k=3 to infinity of 5/k-5/(k+1) ?
The sum from k=3 to infinity of 5/k-5/(k+1) is 5/3