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

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

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

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

converges

Solution steps

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

Take the partial fraction of

\frac{1}{n ^{2} + 5 n} : \frac{1}{5 n} - \frac{1}{5 ( n + 5 )}

= n = 1 \sum \infty \frac{1}{5 n} - \frac{1}{5 ( n + 5 )}

Multiply by the conjugate of

\frac{1}{5 n} - \frac{1}{5 ( n + 5 )} : \frac{\frac{1}{25 n ^{2}} - \frac{1}{25 n ^{2} + 250 n + 625}}{\frac{1}{5 n} + \frac{1}{5 ( n + 5 )}}

= n = 1 \sum \infty \frac{\frac{1}{25 n ^{2}} - \frac{1}{25 n ^{2} + 250 n + 625}}{\frac{1}{5 n} + \frac{1}{5 ( n + 5 )}}

Simplify

\frac{\frac{1}{25 n ^{2}} - \frac{1}{25 n ^{2} + 250 n + 625}}{\frac{1}{5 n} + \frac{1}{5 ( n + 5 )}} : \frac{1}{n ( n + 5 )}

= n = 1 \sum \infty \frac{1}{n ( n + 5 )}

Apply Series Comparison Test:

converges

= converges

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