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

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

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

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

converges

Solution steps

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

Apply Series Root Test:

converges

= converges

n = 1 \sum \infty (\frac{π}{e})^{n}

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

n = 1 \sum \infty \frac{n + 5}{n + 4}

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

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

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