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

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

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

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

converges

Solution steps

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

Apply Series Comparison Test:

converges

= converges

k = 0 \sum \infty (- \frac{1}{2})^{k}

n = 1 \sum \infty \frac{1}{4 n + 4}

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

k = 0 \sum \infty (\frac{3}{4})^{k}

n = 1 \sum \infty (1 - \frac{2}{n})^{3 n}

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