What is the sum from n=3 to infinity of 1/(n^2+3n) ?

The sum from n=3 to infinity of 1/(n^2+3n) is converges

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sum from n=3 to infinity of 1/(n^2+3n)

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

converges

Solution steps

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

Apply Series Comparison Test:

converges

= converges

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x \to - 1 lim (\frac{x}{x ^{2} + x})

f (x) = e^{x^{2} - 6}

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

\frac{\partial}{\partial x} (2 sin (2 x) cos (2 y))

What is the sum from n=3 to infinity of 1/(n^2+3n) ?
The sum from n=3 to infinity of 1/(n^2+3n) is converges