What is the sum from n=0 to infinity of e^{-n^8} ?

The sum from n=0 to infinity of e^{-n^8} is converges

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sum from n=0 to infinity of e^{-n^8}

n = 0 \sum \infty e^{- n^{8}}

converges

Solution steps

n = 0 \sum \infty e^{- n^{8}}

n \to \infty lim (n (\frac{e ^{- n^{8}}}{e ^{- (n + 1)^{8}}} - 1)) = \infty

L > 1,

by the Raabe's test

= converges

n = 2 \sum \infty \frac{4}{n ^{3}}

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

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

n = 0 \sum \infty \frac{3}{8 ^{n}}

n = 0 \sum \infty \frac{n ^{3}}{4 ^{n}}

What is the sum from n=0 to infinity of e^{-n^8} ?
The sum from n=0 to infinity of e^{-n^8} is converges