What is the sum from n=1 to infinity of 4/(n(n+2)) ?

The sum from n=1 to infinity of 4/(n(n+2)) is 3

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sum from n=1 to infinity of 4/(n(n+2))

n = 1 \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

x^{\frac{6}{5}}

y^{^{''}} - 4 y^{^{'}} - 5 y = 9 e^{2 t}

\int - π cos (π) x d x

\frac{d ^{2}}{d x ^{2}} (3 x^{2} e^{x})

\int sec (2)

What is the sum from n=1 to infinity of 4/(n(n+2)) ?
The sum from n=1 to infinity of 4/(n(n+2)) is 3