What is the sum from n=3 to infinity of 5/(n(n+1)) ?

The sum from n=3 to infinity of 5/(n(n+1)) is 5/3

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sum from n=3 to infinity of 5/(n(n+1))

n = 3 \sum \infty \frac{5}{n ( n + 1 )}

\frac{5}{3}

Solution steps

n = 3 \sum \infty \frac{5}{n ( n + 1 )}

Apply the constant multiplication rule:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

= 5 \cdot n = 3 \sum \infty \frac{1}{n ( n + 1 )}

Apply Telescoping Series Test:

\frac{1}{3}

= 5 \cdot \frac{1}{3}

Simplify

5 \cdot \frac{1}{3} : \frac{5}{3}

= \frac{5}{3}

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

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

n = 0 \sum \infty \frac{2}{9 - n}

n = 1 \sum \infty 3 (0.5)^{n - 1}

x = 1 \sum \infty \frac{1}{x}

What is the sum from n=3 to infinity of 5/(n(n+1)) ?
The sum from n=3 to infinity of 5/(n(n+1)) is 5/3