What is the sum from n=1 to infinity of 5/(n^2+7n) ?

The sum from n=1 to infinity of 5/(n^2+7n) is converges

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sum from n=1 to infinity of 5/(n^2+7n)

n = 1 \sum \infty \frac{5}{n ^{2} + 7 n}

converges

Solution steps

n = 1 \sum \infty \frac{5}{n ^{2} + 7 n}

Apply the constant multiplication rule:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

= 5 \cdot 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 )}

= 5 \cdot 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 )}}

= 5 \cdot 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 )}

= 5 \cdot n = 1 \sum \infty \frac{1}{n ( n + 7 )}

Apply Series Comparison Test:

converges

= 5 converges

= converges

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What is the sum from n=1 to infinity of 5/(n^2+7n) ?
The sum from n=1 to infinity of 5/(n^2+7n) is converges