What is the sum from n=5 to infinity of 4/(n^2-n) ?

The sum from n=5 to infinity of 4/(n^2-n) is 1

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sum from n=5 to infinity of 4/(n^2-n)

n = 5 \sum \infty \frac{4}{n ^{2} - n}

1

Solution steps

n = 5 \sum \infty \frac{4}{n ^{2} - n}

Apply the constant multiplication rule:

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

= 4 \cdot n = 5 \sum \infty \frac{1}{n ^{2} - n}

Apply Telescoping Series Test:

\frac{1}{4}

= 4 \cdot \frac{1}{4}

Simplify

4 \cdot \frac{1}{4} : 1

= 1

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

n = 7 \sum \infty \frac{1}{n} - \frac{1}{n + 1}

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

n = 0 \sum \infty \frac{1}{8 - n}

n = 1 \sum \infty 9 (0.1)^{n - 1}

What is the sum from n=5 to infinity of 4/(n^2-n) ?
The sum from n=5 to infinity of 4/(n^2-n) is 1