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

The sum from n=1 to infinity of 3/((n(n+2))) is 9/4

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

n = 1 \sum \infty \frac{3}{( n ( n + 2 ) )}

\frac{9}{4}

Solution steps

n = 1 \sum \infty \frac{3}{n ( n + 2 )}

Apply the constant multiplication rule:

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

= 3 \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 )}

= 3 \cdot n = 1 \sum \infty \frac{1}{2 n} - \frac{1}{2 ( n + 2 )}

Expand telescoping series:

\frac{3}{4}

Simplify

3 \cdot \frac{3}{4} : \frac{9}{4}

= \frac{9}{4}

\int 24 x e^{2.4 x} d x

\frac{d}{d x} (x + 2 x)

\frac{\partial}{\partial x} (x e^{- x^{2}})

y = sec (x) - 2 cos (x), (\frac{π}{3}, 1)

n = 0 \sum \infty (0.7)^{n}

What is the sum from n=1 to infinity of 3/((n(n+2))) ?
The sum from n=1 to infinity of 3/((n(n+2))) is 9/4