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

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

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

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

converges

Solution steps

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

Expand

n (n + 3) : n^{2} + 3 n

= 3 \cdot n = 2 \sum \infty \frac{1}{n ^{2} + 3 n}

Apply Series Comparison Test:

converges

= 3 converges

= converges

k = 1 \sum \infty (cos (6))^{k}

n = 1 \sum \infty \frac{n !}{8 6 ^{n}}

n = 2 \sum \infty \frac{n}{ln ( 9 n )}

k = 1 \sum \infty k (\frac{7}{8})^{k}

n = 1 \sum \infty \frac{5}{2 n ^{2}}

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