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

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

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

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

\frac{2}{3}

Solution steps

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

Apply the constant multiplication rule:

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

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

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 = 0 \sum \infty ln (1 + n^{6})

n = 1 \sum \infty \frac{1}{10 ln ( n )}

n = 0 \sum \infty \frac{( - 5 ) ^{n}}{n}

n = 0 \sum \infty \frac{( 6 ^{n} )}{n !}

n = 0 \sum \infty ln (3 - n)

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