What is the sum from n=1 to infinity of 1/(n^2+7n) ?

The sum from n=1 to infinity of 1/(n^2+7n) is converges

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sum from n=1 to infinity of 1/(n^2+7n)

n = 1 \sum \infty \frac{1}{n ^{2} + 7 n}

converges

Solution steps

n = 1 \sum \infty \frac{1}{n ^{2} + 7 n}

Take the partial fraction of

\frac{1}{n ^{2} + 7 n} : \frac{1}{7 n} - \frac{1}{7 ( n + 7 )}

= n = 1 \sum \infty \frac{1}{7 n} - \frac{1}{7 ( n + 7 )}

Multiply by the conjugate of

\frac{1}{7 n} - \frac{1}{7 ( n + 7 )} : \frac{\frac{1}{49 n ^{2}} - \frac{1}{49 n ^{2} + 686 n + 2401}}{\frac{1}{7 n} + \frac{1}{7 ( n + 7 )}}

= n = 1 \sum \infty \frac{\frac{1}{49 n ^{2}} - \frac{1}{49 n ^{2} + 686 n + 2401}}{\frac{1}{7 n} + \frac{1}{7 ( n + 7 )}}

Simplify

\frac{\frac{1}{49 n ^{2}} - \frac{1}{49 n ^{2} + 686 n + 2401}}{\frac{1}{7 n} + \frac{1}{7 ( n + 7 )}} : \frac{1}{n ( n + 7 )}

= n = 1 \sum \infty \frac{1}{n ( n + 7 )}

Apply Series Comparison Test:

converges

= converges

F (x) = (x^{4} + 9 x^{2} - 7)^{8}

\int_{3}^{9} \frac{8}{x} d x

x \to 4 lim (\frac{7 + x}{4 - x})

x \to 0 lim (\frac{x}{sin ^{2} ( 0 )})

\int \frac{tan ( 2 x )}{2} d x

What is the sum from n=1 to infinity of 1/(n^2+7n) ?
The sum from n=1 to infinity of 1/(n^2+7n) is converges