What is the sum from n=2 to infinity of 7/(n(n+1)) ?

The sum from n=2 to infinity of 7/(n(n+1)) is 7/2

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sum from n=2 to infinity of 7/(n(n+1))

n = 2 \sum \infty \frac{7}{n ( n + 1 )}

\frac{7}{2}

Decimal

3.5

Solution steps

n = 2 \sum \infty \frac{7}{n ( n + 1 )}

Apply the constant multiplication rule:

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

= 7 \cdot n = 2 \sum \infty \frac{1}{n ( n + 1 )}

Apply Telescoping Series Test:

\frac{1}{2}

= 7 \cdot \frac{1}{2}

Simplify

7 \cdot \frac{1}{2} : \frac{7}{2}

= \frac{7}{2}

What is the sum from n=2 to infinity of 7/(n(n+1)) ?
The sum from n=2 to infinity of 7/(n(n+1)) is 7/2