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

x \to - 4 lim (x^{2} - 16)

\frac{d}{d x} (\frac{x ^{\frac{1}{2}}}{3 x})

f (x) = x^{3} + 7 e^{x}

\int e^{3 x} (1 - e^{3 x})^{2} d x

\frac{d}{d x} (\frac{e ^{x}}{x ^{2} + 1})

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