What is the sum from n=8 to infinity of 1/(n^2+64) ?

The sum from n=8 to infinity of 1/(n^2+64) is converges

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sum from n=8 to infinity of 1/(n^2+64)

n = 8 \sum \infty \frac{1}{n ^{2} + 64}

converges

Solution steps

n = 8 \sum \infty \frac{1}{n ^{2} + 64}

Apply Series Comparison Test:

converges

= converges

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

n = 1 \sum \infty \frac{n e ^{n}}{n !}

n = 1 \sum \infty \frac{n}{3 n}

n = 0 \sum \infty \frac{n}{2 n ^{3} + 1}

n = 0 \sum \infty n

What is the sum from n=8 to infinity of 1/(n^2+64) ?
The sum from n=8 to infinity of 1/(n^2+64) is converges