What is the sum from n=1 to infinity of 5/(n^2+n) ?

The sum from n=1 to infinity of 5/(n^2+n) is 5

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sum from n=1 to infinity of 5/(n^2+n)

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

5

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Telescoping Series Test:

1

= 5 \cdot 1

Simplify

= 5

n = 0 \sum \infty \frac{2}{n} - \frac{2}{n + 1}

n = 1 \sum \infty (- 9)^{- n}

k = 0 \sum \infty (\frac{1}{5})^{k}

n = 1 \sum \infty \frac{1}{3 n ^{4} + 4}

n = 3 \sum \infty \frac{n}{2 ^{n}}

What is the sum from n=1 to infinity of 5/(n^2+n) ?
The sum from n=1 to infinity of 5/(n^2+n) is 5