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

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

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

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

n = 0 \sum \infty n^{- s i n (1)}

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

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

n = 2 \sum \infty \frac{7}{n ( n + 1 )}

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