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

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

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

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

diverges

Solution steps

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

Apply Series Root Test:

diverges

= diverges

n = 0 \sum \infty sin (\frac{πn}{4})

n = 1 \sum \infty cosh (in)

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

n = 1 \sum \infty (0.3)^{n}

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

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