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

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

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

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

converges

Solution steps

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

Expand

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

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

Apply Series Comparison Test:

converges

= converges

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\frac{\partial}{\partial x} (2 x y^{3} + cos (y))

\frac{\partial}{\partial x} (e^{y} e^{x} + ln (x y))

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