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$$\sum_{n=0}^\infty n!x^{n^2}$$

Can anyone tell me what will be the radius of convergence for this power series. According to me it is $0$.

If you can please show me how to go about it if I am wrong.

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    How did you arrive at $0$? – 2017-02-18
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    By using the method of |U_(n+1)/U_(n)|. – 2017-02-18
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    It ends up in the form 1/ n^(1/(2n+1)) as n tends to infinity . I am not clear what this value will tend to. But I have assumed it tends to 0 as n tends to infinity – 2017-02-18
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    This ?$$\frac{1}{ n^{\frac{1}{2n+1}}}$$ – 2017-02-18
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    Actually it should be (n+1) instead of n in the denominator rest is OK – 2017-02-18
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    It is $0$. See : Hadamard Radius Formula. – 2017-02-18
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    @user254665 in the formula you have mentioned the power of the exponent is of first degree and here our question has the exponent in second degree ,so I don't think hadamard formula is applicable here. – 2017-02-19
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    The radius is $1$.(I miscalculated). And Hadamard's formula IS applicable. When $a_m=0$ when $m>1$ is not a square, and $a_m=(sqrt m)!$ when $m$ is a square. Then $\lim_{n\to \infty} \sup{j>n}|a_j|^{1/j}=$ $\lim_{m\to \infty}n!^{1/n^2}=1$ by Stirling's Formula. – 2017-02-19

3 Answers 3

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Let's treat this power series as a 'series'.

We will be using root test here.

Let $x$ be fixed. According to the root test, For the convergence of given series we need to assure that $\lim_{n \to \infty} |n! x^{n^2}|^{\frac 1n} \lt 1.$ That is, for sufficiently large values of $n$, $\; |n! x^{n^2}|^{\frac 1n} \lt 1 \Rightarrow |n!|^{\frac 1n}|x^n| \lt 1 \Rightarrow |x^n| \lt |n!|^{-\frac 1n} \Rightarrow |x| \lt |n!|^{-\frac 1{n^2}}$.

Hence $|x| \lt \lim_{n \to \infty} |n!|^{-\frac 1{n^2}}=1$.

Hence the series converges for $|x| \lt 1$. i.e. Radius of convergence is $1$.

EDIT 1: The root test says that ''If $\lim |a_n|^{\frac 1n} \lt 1$, then $\sum a_n$ is absolutely convergent.'' Our $a_n$ here is $n!x^{n^2}$.

EDIT 2: For evaluating $\lim_{n \to \infty} |n!|^{-\frac 1{n^2}}$,

Take $y_n={n!}^{- \frac 1{n^2}} \Rightarrow \ln y_n = -\frac {\ln n!}{n^2}= \frac {\ln \frac 1{n!}}{n^2} \; .\;. \;. \;. \;. \;. \;. \; (1)$

Now $\frac {n \ln \frac 1n}{n^2} \lt \ln y_n \lt \frac {n \ln1}{n^2}$.

Since left side inequality and right side inequalities $\to 0$ as $n \to \infty$,hence by sandwitch thoeorem, $\lim \; \ln y_n =\ln y=0 \Rightarrow y=1$. $\; (\text{Let} \; y_n \to y).$

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    Can you show the procedure? – 2017-02-18
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    The actual limit is different. – 2017-02-18
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    It should be $\lim_{n\to\infty} \left(\frac{1}{n+1}\right)^\frac{1}{2n+1}$. – 2017-02-18
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    @SchrodingersCat I am using root test for convergence of a series $\sum a_n$. The test says that ''if $\lim |a_n|^{\frac 1n} \lt 1$, then $\sum a_n$ is absolutely convergent.'' My $a_n$ here is $n! x^{n^2}$ – 2017-02-18
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    Ok thats fine. I know about the root test. The limit I just gave was the radius of convergence. – 2017-02-18
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    @SchrodingersCat That's why I mentioned very first that let's look at this power series as a series. :) Then root test does the job as you can see from wolfram alpha links that I have showed you. – 2017-02-18
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    @SchrodingersCat So radius of convergence turns out to be $1$. – 2017-02-18
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    How does the ROC come to be 1? Show me that part. You just showed me the root test one. – 2017-02-18
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    @SchrodingersCat Okay root test does shows us $|x| \lt 1$. By definition of radius of convergence, $R$ is radius of convergence of a power series if the series is absolutely convergent on $|x| \lt R$. – 2017-02-18
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    Note that $a_n$ is independent of $x$. So the edit that you have written is wrong. – 2017-02-18
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    Let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/53849/discussion-between-vikrant-desai-and-schrodingerscat). – 2017-02-18
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    @SchrodingersCat For every $-1 \lt x \lt 1$, $\lim_{n \to \infty} a_n=0$. Again note that I've taken $a_n=n! x^{n^2}$. – 2017-02-18
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    @Vikrant Desai a_n is independent of variable x either you say u_n or else its wrong. – 2017-02-18
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    @shadowkh Okay. Then take $u_n= n! x^{n^2}$ and replace it wherever $a_n$ is. – 2017-02-18
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    In fact I like the way you have explained the answer. The right answer is indeed 1. – 2017-02-18
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    @shadowkh I have made EDIT 2 look more easy by sandwich theorem. Thanks. :) I also learned many things that I didn't know before while solving this problem. :) – 2017-02-18
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The radius of convergence is given by $R^{-1}=\mathbb{lim\ sup_{n\rightarrow\infty}} |a_{n}|^{1/n}=\mathbb{lim\ sup_{n\rightarrow\infty}} n!^{1/n^2}=1$.

Edit: I don't know all the forms of the ratio test, but the other answers compute a limit that, in this case, does not exist because we have "holes" in the power serie (the only coefficients that appears are of the form $x^{n^2}$)

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    My edit was "ratio" for "radio". BTW the lim sup IS a limit. It's $\lim_{n\to \infty} b_n$ where $b_n=\sup \{|a_m|^{1/m}:m>n\}$..............+1 – 2017-02-19
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Radius of convergence, R is given by: $$R=\lim_{n\to\infty}\left|\frac{a_n}{a_{n+1}}\right|$$ $$=\lim_{n\to\infty}\left|\frac{n!}{(n+1)!}\right|^\frac{1}{2n+1}$$ $$=\lim_{n\to\infty}\left|\frac{1}{n+1}\right|^\frac{1}{2n+1}$$

Hope this helps you.

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    Yes sure I was expecting this ans – 2017-02-18
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    You're right. They will not be closed intervals but open intervals. Thanks. – 2017-02-18
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    This is a power series in $x$, so the radius of convergence should not depend on $x$. – 2017-02-18
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    Oh yes. Dont know what I was thinking. Correcting it right now. – 2017-02-18
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    @JohnGowers I have corrected it. Thanks for your help. – 2017-02-18
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    I think you calculated radius of convergence for $\sum n! x^n$. Have a look at my answer. :) – 2017-02-18
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    @SchrodingersCat With the edit, you are no longer committing the error of computing the radius of convergence for $\sum n! x^n$, but the exponent $1/(2n+1)$ is very poorly explained. It is non-sensical to compute the limit of $a_n/a_{n+1}$ when $a_n = 0$ infinitely often (I am making an assumption about how you define $a_n$, but this is another thing you have left unexplained). – 2017-02-18
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    I stand corrected the answer is 1 and I have found where I went wrong – 2017-02-18
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    This question was from JAM 2017 Mathematics paper.Isn't it? – 2017-02-19
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    Yes it was and it is given as 1 by many institution – 2017-02-21