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

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

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

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

2

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Telescoping Series Test:

1

= 2 \cdot 1

Simplify

= 2

\frac{d}{d x} (\frac{x ^{3} + 14 x - 7}{7})

\int \frac{ln ( x ^{21} )}{x} d x

\int_{- 2}^{1} (2 x + π) d x

\int \frac{2 + x}{x ^{3}} d x

n = 1 \sum \infty n^{3} e^{- n}

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