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Find the radius of convergence of: $$\sum \frac{1}{n^n\cdot 2^{2n}}(x+2)^{n^2}$$

Please help me in finding the radius of convergence of this power series. I found that it is $\infty.$ But the answer is one. Is my answer correct?

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    Can you show your work? – 2017-01-28
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    It came in my exam today. I submitted my work there. I just wanted to know am I correct? – 2017-01-28
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    No. Take for instance the series $\sum_n \frac{4^{n^2}}{n^n 2^{2n}}$ ($x=2$). Since the general term is $2^{2n^2 - n\log_2 n - 2n} = 2^{2n^2+o(n^2)} \xrightarrow[n\to\infty]{} \infty$, the series diverge (its general term does not even converge to $0$). – 2017-01-28

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HINT:

Using the Root Test, we see that

$$\lim_{n\to \infty}\sqrt[n]{\left|\frac{(x+2)^{n^2}}{n^n2^{2n}}\right|}=\lim_{n\to \infty}\frac{|x+2|^n}{4n} \tag 1$$

For what values of $x$ does the limit in $(1)$ less than $1$?

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    Dr. MV this is the nth root test not ratio test, just for correction. – 2017-01-28
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    @JosephQuarcoo Oops. Root test. I've edited. Thank you Joseph! -Mark – 2017-01-28
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    Actually this is a power series, where the terms are $0$ except when the power is the square of an integer. So if you wanted to, you could take the $1/n^2$ root. Gives the same answer of course. – 2017-01-28
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    @zhw. Yes, I suppose it is since $n^2$ is an integer. I've deleted the preamble. – 2017-01-28
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    @dr mv but what is radius of convergence in your is the value of x? As I am beginner I don't know please help.. – 2017-01-29
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    @AbhishekChandra Note that when $|x+2|\le 1$, the limit on the right-hand side of $(1)$ vanishes (if it's less than $1$, then the series converges). Hence the interval of convergence is $-3\le x\le -1$. If $|x+2|>1$, then the limit is $\infty$ and the series converges. – 2017-02-02
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    Please let me know how I can improve my answer. I really want to give you the best answer I can. -Mark – 2017-02-15