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

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

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

n = 4 \sum \infty \frac{3}{n ( n - 1 )}

1

Solution steps

n = 4 \sum \infty \frac{3}{n ( n - 1 )}

Apply the constant multiplication rule:

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

= 3 \cdot n = 4 \sum \infty \frac{1}{n ( n - 1 )}

Apply Telescoping Series Test:

\frac{1}{3}

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

Simplify

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

= 1

n = 2 \sum \infty \frac{1}{n ^{2} ln ( n )}

i = 1 \sum \infty \frac{1}{3 ^{i}}

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

k = 1 \sum \infty \frac{1}{k ^{2}}

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

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