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Well it seems impossible...

Does this limit even exist (in R)?

I calculated this limit with mathematica and I got $e^{2i\space0\space to\space \pi}$ but I don't know what that is...

How can I tell $\sum_{k=1}^{\infty} \frac{(-1)^k}{k^{\frac{1}{k}}}$ does not converge?

Thanks.

homiee
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2 Answers2

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Hint by $k^{1/k}=e^{\frac{1}{k}\log k}$ we find that $\lim_{k\to\infty}k^{1/k}=1$ and the sequence $((-1)^k)$ has not a limit, so what you can conclude?

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Note that Mathematica says here, that no limit exists. The output 

Exp[2 I Intervall[{0,Pi}]]

Just say that all values that are taken for big enough $k$ are in the form $$\exp(2i x)$$ with $x\in[0,\pi]$. Mathematica thinks $k\in \mathbb{R}$, that is why it gives an Intervall, for $k\in \mathbb{N}$ the sequence has the two accumulation points $-1$ and $1$.

Because $k^\frac{1}{k}$ is converging to $1$, and always greater equal 1, it gives this result.

For the series use that it necessary that the sequence that is summated must be a to $0$ converging sequence.

Dominic Michaelis
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  • Thanks for your answer. What do you mean by "must be a to 0 converging sequence"? Does it mean that the sequence must converge to 0? – homiee Mar 12 '13 at 14:45
  • Lets say $s_n:= \sum_{k=0}^n$ converges to $L$. Than obviously $s_{n-1}$ converges to $L$ too. Than $s_n -s_{n-1}$ converges to zero, and $s_n-s_{n-1}=a_{n}$ – Dominic Michaelis Mar 12 '13 at 14:48