What is the sum from k=3 to infinity of 1/(k^2+5k) ?

The sum from k=3 to infinity of 1/(k^2+5k) is converges

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sum from k=3 to infinity of 1/(k^2+5k)

k = 3 \sum \infty \frac{1}{k ^{2} + 5 k}

converges

Solution steps

k = 3 \sum \infty \frac{1}{k ^{2} + 5 k}

Apply Series Comparison Test:

converges

= converges

n = 1 \sum \infty \frac{9}{- 2 ^{n}}

n = 1 \sum \infty 45 (- 1)^{n}

n = 1 \sum \infty \frac{n ^{6}}{n ^{7}}

n = 1 \sum \infty \frac{4 n}{3 n + 1}

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

What is the sum from k=3 to infinity of 1/(k^2+5k) ?
The sum from k=3 to infinity of 1/(k^2+5k) is converges