What is the sum from n=1 to infinity of e^{-3/(n^2)} ?

The sum from n=1 to infinity of e^{-3/(n^2)} is diverges

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sum from n=1 to infinity of e^{-3/(n^2)}

n = 1 \sum \infty e^{- \frac{3}{n ^{2}}}

diverges

Solution steps

n = 1 \sum \infty e^{- \frac{3}{n ^{2}}}

Apply Series Divergence Test:

diverges

= diverges

\int_{1}^{\infty} \frac{3}{x ^{2} + 1} d x

\int \frac{( 2 x + 1 )}{x} d x

n = 8 \sum \infty (\frac{7}{8})^{- n}

\frac{\partial}{\partial z} (x^{3} + y z^{2})

\frac{\partial}{\partial x} (cos (\frac{x}{1 + y}))

What is the sum from n=1 to infinity of e^{-3/(n^2)} ?
The sum from n=1 to infinity of e^{-3/(n^2)} is diverges