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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 $a=0$.2012-04-20
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    Hint: What is $\lim_{k\to\infty} a^{1/k}$?2012-04-20
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    @Jim Conant: in fact I don't know if it's possible to define $\lim_{k\to\infty}a^{\frac{1}{k}}$2012-04-20
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    @Raskolnikov: why?2012-04-20
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    If you're not sure about the convergence of $\lim_{k\to\infty}{a^{1/k}}$, try calculating some examples and see if you can guess whether the limit exists and if so what it is.2012-04-20
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    Maybe it would help to rewrite it as $e^{\frac{1}{k} \cdot \text{ln} a}$.2012-04-20
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    Is it for any $a\in\mathbb{R}$, or do you also assume that $a > 0$? I guess it doesn't matter in the end, but the case $a < 0$ has to be handled a bit more carefully when computing $\lim_{k \rightarrow \infty} a^{\frac{1}{k}}$.2012-04-20
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    @Nicholas Stull: maybe we can consider only the case $a\ge 0$2012-04-20
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    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.)2012-04-20

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