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Another Question on Radius of convergence :

Calculate Radius and domain of convergence for $$\sum _{n=1}^{\infty}2^n x^{n^2}$$

I used the formula $\lim_{n\rightarrow \infty} |\frac{a_n}{a_{n+1}}|$ to calculate radius of convergence.

I have $\lim_{n\rightarrow \infty} |\frac{a_n}{a_{n+1}}|=\lim_{n\rightarrow \infty} |\frac{2^n}{2^{n+1}}|=\frac{1}{2}$

So,radius of convergence is $\frac{1}{2}$ for $\sum _{n=1}^{\infty}2^n x^n$ but how would i use this to calculate radius of convergence for $$\sum _{n=1}^{\infty}2^n x^{n^2}$$

If i have series something like $\sum _{n=1}^{\infty}2^n (x^n)^2$

I would have just write radius of convergence is $\sqrt{\frac{1}{2}}=\frac{1}{\sqrt{2}}$

But in this case i do not know what to do.. please help me to see this.

Another point is I some how tried to use : For $\sum_{n=1}^{\infty}a_nx^n$ radius would be $R$ where

$$\frac{1}{R}=\lim \sup \sqrt[n]{|a_n|}$$

My book says that if I have some thing like $$\sum _{n=1}^{\infty}a_n x^{n^2}$$ I could use same formula with $$\frac{1}{R}=\lim \sup \sqrt[n^2]{|a_n|}$$

I am not very sure if i can really use it but if i use that, I would get :

$$\frac{1}{R}=\lim \sup \sqrt[n^2]{2^n}=\sqrt{2}$$

SO, radius of convergence would be $\frac{1}{\sqrt{2}}$

which is not much different from what i have got with previous case of $\sum _{n=1}^{\infty}2^n (x^n)^2$

I would like to know more about something similar to this..

Even if i have $\sum _{n=1}^{\infty}a_n x^{n^2}$ or $\sum _{n=1}^{\infty}a_n x^n$, the radius of convergenc is not changing.

Please help me to look into this in more detail.

Thank you

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  • $\begingroup$ Square by $n$ and do the root test perhaps? $\endgroup$ – Don Larynx Dec 20 '13 at 6:17
  • $\begingroup$ I am sorry, I could not get you.... what are you saying? It is true/It is false.. Could you please be more precise... $\endgroup$ – user87543 Dec 20 '13 at 6:19
  • $\begingroup$ You say $\sum _{n=1}^{\infty}2^n (x^n)^2$ I would have just write radius of convergence is $\sqrt{\frac{1}{2}}=\frac{1}{\sqrt{2}}$. Just square it by $n$ $\endgroup$ – Don Larynx Dec 20 '13 at 6:21
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    $\begingroup$ what is "square it by n " I could not understand.. $\endgroup$ – user87543 Dec 20 '13 at 6:27
  • $\begingroup$ @DonLarynx The phrase "square it by $n$" is nonsensical... $\endgroup$ – Potato Dec 20 '13 at 6:52
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Since your series has the form $$\sum_{n}a_n x^{\color{red}{n^2}}$$ and not $$\sum_{n}a_n x^{\color{green}{n}}$$ hence we use the ratio test in this way: let $u_n=a_n x^{n^2}$ $$\lim_{n\to\infty}\left|\frac{u_{n+1}}{u_n}\right|=\lim_{n\to\infty}2|x|^{2n+1}<1\iff |x|<1 $$ hence the radius is $R=1$. Do the same thing for using the Cauchy test.

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    $\begingroup$ Ahhhh, just my style: using color! +1 $\endgroup$ – amWhy Dec 20 '13 at 13:23
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Why not use this, which I guess you mentioned in the question?

You'll have $$C=\lim\sup\sqrt[n]{|2^n\cdot(x^n)^n|}=\lim\sup(2\cdot|x|^n).$$

If $|x|\lt1$, then $C=0\lt 1$. If $|x|\ge1$, then $C\gt1$.

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  • $\begingroup$ Thank you.. This is so helpful :) $\endgroup$ – user87543 Dec 20 '13 at 7:25
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The radius of convergence is $1,$ since obviously the series diverges for $|x|>1,$ and for $|x|<1,$ there exists an $n(x)$ for which $x^{n(x)} < \frac13,$ so for $n>n(x),$ we have $$ 2^n x^{n^2} < 2^n x^{n(x) n} < (2/3)^n.$$

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  • $\begingroup$ I was sure that series diverges for $|x|>1$ and for $|x|<1$.. Your argument does makes sense to me but i am not sure why did that formula did not work? :O $\endgroup$ – user87543 Dec 20 '13 at 6:26
  • $\begingroup$ Because the $n^2$-th root of $2^n$ is $2^{n/n^2} = 2^{1/n} \neq \sqrt{2}.$ $\endgroup$ – Igor Rivin Dec 20 '13 at 6:28
  • $\begingroup$ But i should take lim sup so i thought I should consider $n=1$ at which supremum occurs so lim sup would be $\sqrt{2}$ $\endgroup$ – user87543 Dec 20 '13 at 7:21

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