What is the sum from k=1 to infinity of 1/(k^{1+1/k)} ?

The sum from k=1 to infinity of 1/(k^{1+1/k)} is diverges

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sum from k=1 to infinity of 1/(k^{1+1/k)}

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

diverges

Solution steps

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

Simplify

k^{1 + \frac{1}{k}} : k k^{\frac{1}{k}}

= k = 1 \sum \infty \frac{1}{k k ^{\frac{1}{k}}}

Apply Series Comparison Test:

diverges

= diverges

f (x) = x^{\frac{5}{2}}

x \to - 1 lim (x (x + 3))

n = 0 \sum \infty (\frac{2}{7})^{n}

y = x + 2, y = 2, y = \frac{1}{x + 1}

y^{^{''}} + 256 y = 64 cos (7 t)

What is the sum from k=1 to infinity of 1/(k^{1+1/k)} ?
The sum from k=1 to infinity of 1/(k^{1+1/k)} is diverges