What is the sum from n=1 to infinity of 5/(n(n+1)) ?

The sum from n=1 to infinity of 5/(n(n+1)) is 5

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sum from n=1 to infinity of 5/(n(n+1))

n = 1 \sum \infty \frac{5}{n ( n + 1 )}

5

Solution steps

n = 1 \sum \infty \frac{5}{n ( n + 1 )}

Apply the constant multiplication rule:

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

= 5 \cdot n = 1 \sum \infty \frac{1}{n ( n + 1 )}

Apply Telescoping Series Test:

1

= 5 \cdot 1

Simplify

= 5

k = 1 \sum \infty \frac{4}{k}

n = 1 \sum \infty n (- \frac{3}{7})^{n}

n = 1 \sum \infty \frac{5}{n ^{2} + 36}

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

k = 1 \sum \infty (cos (10))^{k}

What is the sum from n=1 to infinity of 5/(n(n+1)) ?
The sum from n=1 to infinity of 5/(n(n+1)) is 5