What is the sum from k=2 to infinity of (k^e)/(k^pi) ?

The sum from k=2 to infinity of (k^e)/(k^pi) is diverges

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sum from k=2 to infinity of (k^e)/(k^pi)

k = 2 \sum \infty \frac{k ^{e}}{k ^{π}}

diverges

Solution steps

k = 2 \sum \infty \frac{k ^{e}}{k ^{π}}

Apply Series Integral Test:

diverges

= diverges

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

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}

What is the sum from k=2 to infinity of (k^e)/(k^pi) ?
The sum from k=2 to infinity of (k^e)/(k^pi) is diverges