What is the sum from n=1 to infinity of 4/(n^2+n) ?

The sum from n=1 to infinity of 4/(n^2+n) is 4

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sum from n=1 to infinity of 4/(n^2+n)

n = 1 \sum \infty \frac{4}{n ^{2} + n}

4

Solution steps

n = 1 \sum \infty \frac{4}{n ^{2} + n}

Apply the constant multiplication rule:

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

= 4 \cdot n = 1 \sum \infty \frac{1}{n ^{2} + n}

Apply Telescoping Series Test:

1

= 4 \cdot 1

Simplify

= 4

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

n = 0 \sum \infty \frac{5}{2 n - 3}

n = 2 \sum \infty (1 + \frac{1}{4})^{- n}

n = 1 \sum \infty (- \frac{7}{8})^{n - 1}

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

What is the sum from n=1 to infinity of 4/(n^2+n) ?
The sum from n=1 to infinity of 4/(n^2+n) is 4