What is the sum from n=1 to infinity of e^{-n^2} ?

The sum from n=1 to infinity of e^{-n^2} is converges

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sum from n=1 to infinity of e^{-n^2}

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

converges

Solution steps

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

Apply Series Ratio Test:

converges

= converges

\int e^{\frac{x}{8}} d x

sin (2 x), cos (x), [\frac{π}{6}, \frac{π}{2}]

y = - ln (x)

x \to 0 lim (\frac{sin ( 8 x )}{sin ( 7 x )})

\frac{d}{d x} (x^{2} + y)

What is the sum from n=1 to infinity of e^{-n^2} ?
The sum from n=1 to infinity of e^{-n^2} is converges