What is the sum from n=3 to infinity of 1/(n^2+n) ?

The sum from n=3 to infinity of 1/(n^2+n) is 1/3

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sum from n=3 to infinity of 1/(n^2+n)

n = 3 \sum \infty \frac{1}{n ^{2} + n}

\frac{1}{3}

Solution steps

n = 3 \sum \infty \frac{1}{n ^{2} + n}

Apply Telescoping Series Test:

\frac{1}{3}

= \frac{1}{3}

n = 0 \sum \infty \frac{1}{n ^{0.9}}

n = 0 \sum \infty π

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

n = 1 \sum \infty (\frac{1}{5})^{n} + \frac{3}{n}

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

What is the sum from n=3 to infinity of 1/(n^2+n) ?
The sum from n=3 to infinity of 1/(n^2+n) is 1/3