What is the sum from n=0 to infinity of 4/(n(n+2)) ?

The sum from n=0 to infinity of 4/(n(n+2)) is 3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

sum from n=0 to infinity of 4/(n(n+2))

n = 0 \sum \infty \frac{4}{n ( n + 2 )}

3

Solution steps

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

Apply the constant multiplication rule:

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

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

Take the partial fraction of

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

= 4 \cdot n = 1 \sum \infty \frac{1}{2 n} - \frac{1}{2 ( n + 2 )}

Expand telescoping series:

\frac{3}{4}

Simplify

4 \cdot \frac{3}{4} : 3

= 3

n = 0 \sum \infty \frac{( 5 ^{n} )}{n !}

tan^{2} (θ)

\int \frac{w ^{2}}{( w ^{2} + 7 ) ^{2}} d w

y^{^{'}} = - \frac{2 x}{y}, y (1) = 10

y = (arctan (7 x))^{2}

What is the sum from n=0 to infinity of 4/(n(n+2)) ?
The sum from n=0 to infinity of 4/(n(n+2)) is 3