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Find the convergence radiuses of this power series:

$1 + n + n^{4} + n^{9} + n^{16} + n^{25} + n^{36} + ...$

First of all, I'm surprised it says $radiuses$ instead of $radius$. I know you find the radius by finding the limit of the series aka where it converges and take its reciprocal. But how and is it even possible to find more than just 1 radius for a power series? That would be my first question.

For the task, it doesn't really look like a series because the totals formula is missing, instead we got "...". So I would change it to:

$\sum_{k=0}^{\infty } n^{k^{2}}$

Now we need to find out if this series converges and what its limit point towards $\infty$ is. I don't know what rule to use here, till now we mostly had fractions where you could easily use the ratio test. However, as it looks like this series will go to $\infty$, so it will diverge thus there is no limit point and no convergence radius...? Can't be or maybe I converted it badly to a series at the beginning?

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  • $\begingroup$ $\sum_{k=0}^\infty a_k x^k$ where $a_k=1$ if $k=m^2$ for some $m$ and zero otherwise $\endgroup$ – clark Jun 25 '16 at 21:44
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    $\begingroup$ For $|n|<1$, $|n^{k^2}|<|n^k|$ and the series converges by comparison. And for $n\ge 1$, the series diverges since the terms of the series don't approach $0$. $\endgroup$ – Mark Viola Jun 25 '16 at 21:48
  • $\begingroup$ You can also conclude this from the root test. $\endgroup$ – SquirtleSquad Jun 25 '16 at 21:48
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    $\begingroup$ That's not a power series. $\endgroup$ – zhw. Jun 25 '16 at 22:15
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As @Dr.MV said, we can use the comparison test to solve this problem.

  • $\lvert n \rvert < 1 \rightarrow$ In this case, we have $\lvert n^{k^2} \rvert < \lvert n^k \rvert$, so since $\sum_{k=0}^\infty n^k$ converges, so does $\sum_{k=0}^\infty n^{k^2}$.
  • $\lvert n \rvert \geq 1 \rightarrow$ In this case, we have $\lvert n^{k^2} \rvert \geq \lvert n^k \rvert$, so since $\sum_{k=0}^\infty n^k$ diverges, so does $\sum_{k=0}^\infty n^{k^2}$.

Thus, the sum converges if and only if $\lvert n \rvert < 1$.

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    $\begingroup$ But to what does it converge? There is no specific number? Need that for the radius. Tyvm for clarification btw. $\endgroup$ – kathelk Jun 25 '16 at 21:54
  • $\begingroup$ @kalthelk I don't know what the actual value/formula of convergence is, but since it converges for $\lvert n \rvert < 1$, we know that the radius of convergence is $1$. $\endgroup$ – Noble Mushtak Jun 25 '16 at 21:56
  • $\begingroup$ Ok I think I don't really understand.. : / Anyway, what about my first question? You know if it's possible to get several radiuses for one series? And how would you count them? I only know how to get one radius... $\endgroup$ – kathelk Jun 25 '16 at 21:57
  • $\begingroup$ @kathelk For this series, there is only one radius: The radius of $1$ around the center $n=0$. However, for other series, like $\sum_{k=0}^\infty (2\sin\theta)^k$, there are multiple radii because it converges for $-\frac 1 2 < \sin\theta < \frac 1 2$ which is an inequality that is true for multiple intervals of $\theta$. $\endgroup$ – Noble Mushtak Jun 25 '16 at 21:59
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The series diverges and does not seem to be a power series, so the question in my opinion is badly worded. To see why it diverges, rewrite your series as: $$1+\sum_{n=1}^\infty\frac{1}{n^{-k^2}}.$$ Now using the $p-$series test, for convergence we need $-k^2>1$, which is impossible for all real $k$.

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    $\begingroup$ It does not diverge for $\lvert n \rvert < 1$, though $\endgroup$ – Noble Mushtak Jun 25 '16 at 21:49
  • $\begingroup$ It goes from $k=0$ to $\infty$. $n$ is the variable we are looking at in the power series and $k$ is the index. $\endgroup$ – Noble Mushtak Jun 25 '16 at 21:57
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    $\begingroup$ in OP's question $k^2$ is not fixed, $n$ is fixed in the summation. $\endgroup$ – clark Jun 25 '16 at 21:58
  • $\begingroup$ Yes I see my error. Thanks. $\endgroup$ – John Molokach Jun 26 '16 at 0:23

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