What is the sum from n=1 to infinity of (2^nn)/(n^n) ?

The sum from n=1 to infinity of (2^nn)/(n^n) is converges

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sum from n=1 to infinity of (2^nn)/(n^n)

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

converges

Solution steps

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

n \to \infty lim n \frac{\frac{2 ^{n} n}{n ^{n}}}{\frac{2 ^{n + 1} ( n + 1 )}{( n + 1 ) ^{n + 1}}} - 1 = \infty

L > 1,

by the Raabe's test

= converges

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

n = 0 \sum \infty \frac{6}{1 - n ^{2}}

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

n = 1 \sum \infty 9^{- n}

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

What is the sum from n=1 to infinity of (2^nn)/(n^n) ?
The sum from n=1 to infinity of (2^nn)/(n^n) is converges