What is the sum from n=1 to infinity of n/(n^n) ?

The sum from n=1 to infinity of n/(n^n) is converges

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sum from n=1 to infinity of n/(n^n)

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

converges

Solution steps

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

n \to \infty lim (n (\frac{\frac{n}{n ^{n}}}{\frac{n + 1}{( n + 1 ) ^{n + 1}}} - 1)) = \infty

L > 1,

by the Raabe's test

= converges

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

n = 1 \sum \infty \frac{7}{5 n - 1}

n = 2 \sum \infty (\frac{1}{8})^{n + 4}

n = 0 \sum \infty \frac{3 n}{1 + n ^{4}}

n = 1 \sum \infty 6 - 5 (0.7)^{n}

What is the sum from n=1 to infinity of n/(n^n) ?
The sum from n=1 to infinity of n/(n^n) is converges