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Is it possible to find an analytical expression of the series: $$S=\sum_{k=1}^{N}a^{\frac{1}{k}}$$ where $a$ is a real number? If we have: $$S_{\infty}=\lim_{N\to\infty}S$$ is $S_{\infty}$ convergent for any $a\in\Re$?

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It will not be convergent except for $a=0$. – Raskolnikov Apr 20 '12 at 11:05
Hint: What is $\lim_{k\to\infty} a^{1/k}$? – Grumpy Parsnip Apr 20 '12 at 11:59
@Jim Conant: in fact I don't know if it's possible to define $\lim_{k\to\infty}a^{\frac{1}{k}}$ – Riccardo.Alestra Apr 20 '12 at 12:05
@Raskolnikov: why? – Riccardo.Alestra Apr 20 '12 at 12:06
If $a = 0$, then $a^{1/k} = 0$, $\forall k$. So the series does converge at $a = 0$. Now, let $a > 0$. To compute $\lim_{k\to\infty} a^{1/k}$, find $\lim_{k\to\infty} \ln(a^{1/k}) = L$, and then $\lim_{k\to\infty} a^{1/k} = e^L$. We would then conclude that $\lim_{k\to\infty} a^{1/k} = 1$, $\forall a > 0$. (This result would not be as quick to compute if $a < 0$, but I'm confident it is still true, by expressing $a = |a| e^{i\pi}$, and continuing in a similar manner.) – Nicholas Stull Apr 20 '12 at 14:20

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