What is the sum from n=0 to infinity of 1/(n^2+4) ?

The sum from n=0 to infinity of 1/(n^2+4) is converges

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sum from n=0 to infinity of 1/(n^2+4)

n = 0 \sum \infty \frac{1}{n ^{2} + 4}

converges

Solution steps

n = 0 \sum \infty \frac{1}{n ^{2} + 4}

Apply Series Comparison Test:

converges

= converges

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

n = 1 \sum \infty 4 (\frac{1}{2})^{n - 2}

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

n = 1 \sum \infty \frac{n ^{9}}{9 ^{n}}

n = 1 \sum \infty n^{4} e^{- n^{2}}

What is the sum from n=0 to infinity of 1/(n^2+4) ?
The sum from n=0 to infinity of 1/(n^2+4) is converges