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

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

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

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

1

Solution steps

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

Apply Telescoping Series Test:

1

= 1

\int \frac{x}{x ^{2} - 6 x} d x

y^{^{'}} + y = \frac{1}{1 + e ^{t}}

y^{^{'}} + 7 x^{6} y = x^{6}

x \to 0 lim (x^{2})

x \to 1 lim (6 x)

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