What is the sum from n=0 to infinity of (e)^{-n^2} ?

The sum from n=0 to infinity of (e)^{-n^2} is converges

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sum from n=0 to infinity of (e)^{-n^2}

n = 0 \sum \infty (e)^{- n^{2}}

converges

Solution steps

n = 0 \sum \infty e^{- n^{2}}

Apply Series Ratio Test:

converges

= converges

n = 1 \sum \infty cos (\frac{13}{n})

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

n = 1 \sum \infty (\frac{12}{7})^{n}

n = 1 \sum \infty n (\frac{5}{6})^{n - 1}

n = 4 \sum \infty \frac{1}{n ^{5}}

What is the sum from n=0 to infinity of (e)^{-n^2} ?
The sum from n=0 to infinity of (e)^{-n^2} is converges