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

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

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

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

converges

Solution steps

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

Apply exponent rule:

\frac{a ^{n}}{b ^{n}} = (\frac{a}{b})^{n}

\frac{2 ^{n}}{n ^{n}} = (\frac{2}{n})^{n}

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

Apply Series Root Test:

converges

= converges

n = 0 \sum \infty \frac{n}{10 n + 1}

n = 0 \sum \infty n ln (n)

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

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

n = 1 \sum \infty \frac{4}{7 ^{n}}

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