What is the sum from n=1 to infinity of 5/(n^2+9n+7) ?

The sum from n=1 to infinity of 5/(n^2+9n+7) is converges

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sum from n=1 to infinity of 5/(n^2+9n+7)

n = 1 \sum \infty \frac{5}{n ^{2} + 9 n + 7}

converges

Solution steps

n = 1 \sum \infty \frac{5}{n ^{2} + 9 n + 7}

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} + 9 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+9n+7) ?
The sum from n=1 to infinity of 5/(n^2+9n+7) is converges