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When I was trying to solve this question

I came up to this: $$\lim _{ n\rightarrow \infty }{ \sum _{ k=1 }^{ n }{ { k }^{ \left( \frac { 1 }{ 2k } \right) } } } ={ 1 }^{ 1/2 }+{ 2 }^{ 1/4 }+{ 3 }^{ 1/6 }+...+{ n }^{ 1/(2n) }.$$Does it converge? What is the limit of the sum?

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up vote 3 down vote accepted

This has no limit since $k^{\frac{1}{2k}}\to 1$ and $k\to\infty$.

You don't really need that to see that it can't converge, since $k^{\frac{1}{2k}}>1$ for all $k$.

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Or this has limit $+\infty$. – 1015 May 28 '13 at 16:32

$$\lim_n \sqrt[2n]{n} =1 \neq 0$$

thus the series diverges.

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