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 $a\in\Re$?
Power series calculation
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sequences-and-series
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1If $a = 0$, 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