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

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

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

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

diverges

Solution steps

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

Simplify

\frac{n}{n ^{2} + n} : \frac{1}{n + 1}

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

Apply Series Limit Comparison Test:

diverges

= diverges

n = 0 \sum \infty (\frac{2}{n})^{\frac{5}{n}}

n = 1 \sum \infty \frac{1}{3 n ^{2} + 7}

n = 1 \sum \infty \frac{1}{3 n ^{2} + 6}

n = 1 \sum \infty \frac{32}{n !}

n = 0 \sum \infty \frac{n}{n ^{2} + 7}

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