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

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

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

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

diverges

Solution steps

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

n^{- 2} = \frac{1}{n ^{2}}

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

Apply Series Divergence Test:

diverges

= diverges

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

n = 1 \sum \infty \frac{2}{n}

n = 0 \sum \infty ln (\frac{1}{n})

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

n = 1 \sum \infty 8 - \frac{5}{n ^{2}}

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