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The question is to calculate the radius of convergence of $\sum\limits_{n=0}^\infty n!x^{n^2}$.

I dont know how to calculate R.O.C for such type of questions involving $n^2$ as power of x. How do we do this? Any help appreciated.

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I think the question is to find the radius of convergence, not to "calculate" the series (I doubt that the sum of the series has a closed-form expression).

Ratio test:

$$ \left|\frac{(n+1)!\; x^{(n+1)^2}}{n!\; x^{n^2}}\right| = (n+1) |x|^{2n+1}$$

If $|x| < 1$ this goes to $0$ as $n \to \infty$, and thus is less than $1$ for sufficiently large $n$, thus the series converges.

If $|x| \ge 1$ it is greater than $1$, and the series diverges.

Thus the radius of convergence is $1$.

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    Edited question. Understood the answer. Thank u. – 2017-02-17
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    Sorry for replying so late. I have a doubt. The coefficients are n!, so 1,2,6... and so on. Also we know that 1/R = limsup |a_n|^(1/n), and this limit evaluates to infinity in wolfram alpha, and therefore R.O.C should be 0. Please correct as to where i am wrong. Thank u – 2017-03-21
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    If you're taking $a_n = n!$, you're asking Wolfram Alpha for the R.O.C. of $\sum n! x^n$, not this one. – 2017-03-21
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    But, the series in the question evaluates to $x + 2x^4 + 6x^9 +...$, so it is equivalent to the series of sigma $n!x^n$ just missing some powers of x, whose coefficient is 0. So why are the answers different? Please help – 2017-03-21
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    It is not at all equivalent. For example, in $\sum n! x^n$, the coefficient of $x^9$ is $9! = 362880$, in $\sum n! x^{n^2}$ it is $3! = 6$. – 2017-03-21
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    Sorry i phrased it wrong. I just want to know why cant i use the limsup formula for R.O.C on the series sigma $n!x^(n^2)$. You have said that if i use it, then i am rather evaluating R.O.C of sigma $n!x^n$. Why? Is the formula not applicable for series of the first type? Correct me if i am wrong, but the first series too is the same "type" as series second. Just missing some powers of x, which have coefficients 0. So, why is the limsup formula not applicable here? Help! – 2017-03-22
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    For $k = n^2$, $a_k = n! = (\sqrt{k})!$, so you would need $\limsup_{k \to \infty} ((\sqrt{k})!)^{1/k}$. [The result](http://www.wolframalpha.com/input/?i=limit+of+((sqrt(k))!)%5E(1%2Fk)+as+k+goes+to+infinity) is $1$, as expected. – 2017-03-22
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    Oh ok. Got it. Thank u very much :) – 2017-03-23