What is the sum from n=1 to infinity of 1/(n^2+6n) ?

The sum from n=1 to infinity of 1/(n^2+6n) is converges

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sum from n=1 to infinity of 1/(n^2+6n)

n = 1 \sum \infty \frac{1}{n ^{2} + 6 n}

converges

Solution steps

n = 1 \sum \infty \frac{1}{n ^{2} + 6 n}

Take the partial fraction of

\frac{1}{n ^{2} + 6 n} : \frac{1}{6 n} - \frac{1}{6 ( n + 6 )}

= n = 1 \sum \infty \frac{1}{6 n} - \frac{1}{6 ( n + 6 )}

Multiply by the conjugate of

\frac{1}{6 n} - \frac{1}{6 ( n + 6 )} : \frac{\frac{1}{36 n ^{2}} - \frac{1}{36 n ^{2} + 432 n + 1296}}{\frac{1}{6 n} + \frac{1}{6 ( n + 6 )}}

= n = 1 \sum \infty \frac{\frac{1}{36 n ^{2}} - \frac{1}{36 n ^{2} + 432 n + 1296}}{\frac{1}{6 n} + \frac{1}{6 ( n + 6 )}}

Simplify

\frac{\frac{1}{36 n ^{2}} - \frac{1}{36 n ^{2} + 432 n + 1296}}{\frac{1}{6 n} + \frac{1}{6 ( n + 6 )}} : \frac{1}{n ( n + 6 )}

= n = 1 \sum \infty \frac{1}{n ( n + 6 )}

Apply Series Comparison Test:

converges

= converges

\int x^{9} e^{x} d x

y^{^{''}} + 4 \cdot y = 3 \cdot x - 1

\int_{0}^{0.5} 625 x d x

f (x) = x 3^{x}

What is the sum from n=1 to infinity of 1/(n^2+6n) ?
The sum from n=1 to infinity of 1/(n^2+6n) is converges