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

The sum from n=1 to infinity of 3/((n(n+3))) is 11/6

Popular >

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

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

\frac{11}{6}

Solution steps

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

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 + 3 )}

Take the partial fraction of

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

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

Expand telescoping series:

\frac{11}{18}

Simplify

3 \cdot \frac{11}{18} : \frac{11}{6}

= \frac{11}{6}

k = 1 \sum \infty \frac{9}{k ( k + 2 )}

n = 1 \sum \infty \frac{- 5}{n}

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

n = 1 \sum \infty (\frac{5}{7})^{n - 1}

n = 1 \sum \infty (- \frac{3}{7})^{n}

What is the sum from n=1 to infinity of 3/((n(n+3))) ?
The sum from n=1 to infinity of 3/((n(n+3))) is 11/6