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

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

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

n = 4 \sum \infty \frac{1}{n ^{2} + n}

\frac{1}{4}

Solution steps

n = 4 \sum \infty \frac{1}{n ^{2} + n}

Apply Telescoping Series Test:

\frac{1}{4}

= \frac{1}{4}

x \to 3 lim (\frac{x + 2}{x - 3})

x \to 9 + lim (\frac{x - 9}{∣ x - 9 ∣})

y^{^{''}} + 6 y^{^{'}} + 9 y = 12 e^{- x}

\frac{d}{d x} (cos (\frac{1 - e ^{8 x}}{1 + e ^{8 x}}))

\int \frac{z}{1 0 ^{z}} d z

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