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

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

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sum from n=2 to infinity of 3/(n^2-n)

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

3

Solution steps

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

Apply the constant multiplication rule:

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

= 3 \cdot n = 2 \sum \infty \frac{1}{n ^{2} - n}

Apply Telescoping Series Test:

1

= 3 \cdot 1

Simplify

= 3

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

n = 2 \sum \infty (\frac{3}{7})^{3 n}

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

n = 1 \sum \infty \frac{1}{n ^{25}}

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

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