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

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

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sum from n=1 to infinity of 2/(n*(n+2))

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

\frac{3}{2}

Solution steps

n = 1 \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 = 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 )}

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

Expand telescoping series:

\frac{3}{4}

Simplify

2 \cdot \frac{3}{4} : \frac{3}{2}

= \frac{3}{2}

f (x) = 2 sec (x)

\frac{d}{d x} (80 (1 - 0.1 x)^{3})

\frac{d}{d x} (ln (\frac{x + b}{x}))

y^{^{'}} + 8 y = 5

\int \frac{3 x ^{4} + 4}{x ^{2} ( x ^{2} + 1 ) ^{3}} d x

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