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 \cdot ( 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

\frac{3}{5} x^{\frac{5}{3}} - \frac{3}{4} x^{\frac{1}{3}} + 7

\int (t^{2})^{2} + (t)^{2} d t

n \to 0 lim (\frac{1}{n} sin (\frac{1}{n}))

x \to 0 + lim (- x)

\frac{d}{d x} (\frac{3}{( 4 x - 5 )})

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