What is the sum from n=4 to infinity of 2/(n(n-1)) ?

The sum from n=4 to infinity of 2/(n(n-1)) is 2/3

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sum from n=4 to infinity of 2/(n(n-1))

n = 4 \sum \infty \frac{2}{n ( n - 1 )}

\frac{2}{3}

Solution steps

n = 4 \sum \infty \frac{2}{n ( n - 1 )}

Apply the constant multiplication rule:

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

= 2 \cdot n = 4 \sum \infty \frac{1}{n ( n - 1 )}

Apply Telescoping Series Test:

\frac{1}{3}

= 2 \cdot \frac{1}{3}

Simplify

2 \cdot \frac{1}{3} : \frac{2}{3}

= \frac{2}{3}

n = 1 \sum \infty \frac{n}{n ^{2} + 7}

n = 0 \sum \infty \frac{1}{n \cdot 3 ^{n}}

n = 0 \sum \infty 7

n = 0 \sum \infty - 3 + 4 n^{5}

n = 1 \sum \infty 2 (0.2)^{n - 1}

What is the sum from n=4 to infinity of 2/(n(n-1)) ?
The sum from n=4 to infinity of 2/(n(n-1)) is 2/3