What is the sum from n=1 to infinity of 1/(n^2+16) ?

The sum from n=1 to infinity of 1/(n^2+16) is converges

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sum from n=1 to infinity of 1/(n^2+16)

n = 1 \sum \infty \frac{1}{n ^{2} + 16}

converges

Solution steps

n = 1 \sum \infty \frac{1}{n ^{2} + 16}

Apply Series Comparison Test:

converges

= converges

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

n = 0 \sum \infty (\frac{5}{8})^{n}

n = 0 \sum \infty \frac{e ^{n}}{n ^{3}}

n = 1 \sum \infty (- 6)^{- n}

n = 1 \sum \infty \frac{1}{4 ^{n - 1}}

What is the sum from n=1 to infinity of 1/(n^2+16) ?
The sum from n=1 to infinity of 1/(n^2+16) is converges