What is the sum from n=1 to infinity of 1/((n^2+8n)) ?

The sum from n=1 to infinity of 1/((n^2+8n)) is converges

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sum from n=1 to infinity of 1/((n^2+8n))

n = 1 \sum \infty \frac{1}{( n ^{2} + 8 n )}

converges

Solution steps

n = 1 \sum \infty \frac{1}{n ^{2} + 8 n}

Take the partial fraction of

\frac{1}{n ^{2} + 8 n} : \frac{1}{8 n} - \frac{1}{8 ( n + 8 )}

= n = 1 \sum \infty \frac{1}{8 n} - \frac{1}{8 ( n + 8 )}

Multiply by the conjugate of

\frac{1}{8 n} - \frac{1}{8 ( n + 8 )} : \frac{\frac{1}{64 n ^{2}} - \frac{1}{64 n ^{2} + 1024 n + 4096}}{\frac{1}{8 n} + \frac{1}{8 ( n + 8 )}}

= n = 1 \sum \infty \frac{\frac{1}{64 n ^{2}} - \frac{1}{64 n ^{2} + 1024 n + 4096}}{\frac{1}{8 n} + \frac{1}{8 ( n + 8 )}}

Simplify

\frac{\frac{1}{64 n ^{2}} - \frac{1}{64 n ^{2} + 1024 n + 4096}}{\frac{1}{8 n} + \frac{1}{8 ( n + 8 )}} : \frac{1}{n ( n + 8 )}

= n = 1 \sum \infty \frac{1}{n ( n + 8 )}

Apply Series Comparison Test:

converges

= converges

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What is the sum from n=1 to infinity of 1/((n^2+8n)) ?
The sum from n=1 to infinity of 1/((n^2+8n)) is converges