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

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

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

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

converges

Solution steps

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

Apply Series Root Test:

converges

= converges

n = 0 \sum \infty (4 + n^{2})^{- 1}

k = 0 \sum \infty 1^{k}

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

n = 1 \sum \infty \frac{2 ^{n}}{n ^{4}}

n = 0 \sum \infty \frac{1}{n ^{2} - n}

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