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

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

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

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

diverges

Solution steps

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

Apply Series Integral Test:

diverges

= diverges

n = 1 \sum \infty - 4 (\frac{5}{8})^{n}

k = 1 \sum \infty \frac{9}{k} - \frac{9}{k + 1}

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

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

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

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