What is the sum from n=2 to infinity of 6/(n(n-1)) ?

The sum from n=2 to infinity of 6/(n(n-1)) is 6

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sum from n=2 to infinity of 6/(n(n-1))

n = 2 \sum \infty \frac{6}{n ( n - 1 )}

6

Solution steps

n = 2 \sum \infty \frac{6}{n ( n - 1 )}

Apply the constant multiplication rule:

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

= 6 \cdot n = 2 \sum \infty \frac{1}{n ( n - 1 )}

Apply Telescoping Series Test:

1

= 6 \cdot 1

Simplify

= 6

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

n = 0 \sum \infty (\frac{4}{25})^{n}

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

k = 1 \sum \infty \frac{k}{500 k + 5}

n = 8 \sum \infty \frac{2 n + 1}{n ^{n}}

What is the sum from n=2 to infinity of 6/(n(n-1)) ?
The sum from n=2 to infinity of 6/(n(n-1)) is 6