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

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

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

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

\frac{1}{2}

Solution steps

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

Apply Telescoping Series Test:

\frac{1}{2}

= \frac{1}{2}

\frac{1}{s ^{2} - 3}

\frac{d}{d x} (- 3 e^{1 - 3 x})

\frac{d}{d x} (2 x + 4)

\frac{\partial}{\partial x} (\frac{( - y )}{x ^{2} + y ^{2}})

y^{^{''}} = a y

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