What is the sum from n=5 to infinity of 1/(n^2+n) ?

The sum from n=5 to infinity of 1/(n^2+n) is 1/5

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sum from n=5 to infinity of 1/(n^2+n)

n = 5 \sum \infty \frac{1}{n ^{2} + n}

\frac{1}{5}

Solution steps

n = 5 \sum \infty \frac{1}{n ^{2} + n}

Apply Telescoping Series Test:

\frac{1}{5}

= \frac{1}{5}

\int_{0}^{\frac{π}{3}} cos (4 x) d x

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

y = 8 x e^{x}

\frac{d}{d z} (\frac{2 z ^{6} + 2}{1 z ^{2} + 5})

n = 1 \sum \infty (\frac{e}{n})^{2 n}

What is the sum from n=5 to infinity of 1/(n^2+n) ?
The sum from n=5 to infinity of 1/(n^2+n) is 1/5