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

The sum from n=1 to infinity of 1/(n^2+2n) is 3/4

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

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

\frac{3}{4}

Solution steps

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

Take the partial fraction of

\frac{1}{n ^{2} + 2 n} : \frac{1}{2 n} - \frac{1}{2 ( n + 2 )}

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

Expand telescoping series:

\frac{3}{4}

= \frac{3}{4}

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

f (x) = (1 - x) (x^{2} - 3)^{2}

n = 0 \sum \infty 1^{2 n}

\frac{\partial}{\partial z} (x cos (z))

\frac{d y}{d x} = y (\frac{1}{x} - y)

What is the sum from n=1 to infinity of 1/(n^2+2n) ?
The sum from n=1 to infinity of 1/(n^2+2n) is 3/4