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

The sum from n=1 to infinity of 2/(n(n+3)) is 11/9

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

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

\frac{11}{9}

Solution steps

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

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 + 3 )}

Take the partial fraction of

\frac{1}{n ( n + 3 )} : \frac{1}{3 n} - \frac{1}{3 ( n + 3 )}

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

Expand telescoping series:

\frac{11}{18}

Simplify

2 \cdot \frac{11}{18} : \frac{11}{9}

= \frac{11}{9}

x \to 5 lim (4 x - 3 x^{2})

x \to 5 lim (x^{2} + 2 x - 35)

y = 3 csc^{2} (5 x)^{3}

f (x) = x^{3} + 3 x^{2} - 45 x - 19

y = x^{3} - 2 x + 2 (3.23)

What is the sum from n=1 to infinity of 2/(n(n+3)) ?
The sum from n=1 to infinity of 2/(n(n+3)) is 11/9