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

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

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

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

5

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Telescoping Series Test:

1

= 5 \cdot 1

Simplify

= 5

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

n = 5 \sum \infty \frac{3}{n ^{2} + 5 n + 4}

n = 0 \sum \infty \frac{5 n !}{2 ^{n}}

n = 0 \sum \infty \frac{5}{6 ^{n}}

n = 2 \sum \infty (\frac{3}{8})^{4 n}

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