What is the sum from n=0 to infinity of 1/(n^2+2n) ?

The sum from n=0 to infinity of 1/(n^2+2n) is 3/4

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sum from n=0 to infinity of 1/(n^2+2n)

n = 0 \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}

\int \frac{5000 x}{25 x + 1500} d x

x \to 0 lim (\frac{1}{4 x})

\frac{d}{d x} (cos (2 x + 5))

\frac{d}{d x} (\frac{5 x ^{4} - 3 x ^{3}}{x ^{5}})

\int \frac{1}{4} (2 x + 1) d x

What is the sum from n=0 to infinity of 1/(n^2+2n) ?
The sum from n=0 to infinity of 1/(n^2+2n) is 3/4