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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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  • $\begingroup$ Can you show your work? $\endgroup$ – Clement C. Jan 28 '17 at 20:52
  • $\begingroup$ It came in my exam today. I submitted my work there. I just wanted to know am I correct? $\endgroup$ – Abhishek Chandra Jan 28 '17 at 20:53
  • $\begingroup$ 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 $0$). $\endgroup$ – Clement C. Jan 28 '17 at 20:56
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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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  • $\begingroup$ Dr. MV this is the nth root test not ratio test, just for correction. $\endgroup$ – Joseph Quarcoo Jan 28 '17 at 21:11
  • $\begingroup$ @JosephQuarcoo Oops. Root test. I've edited. Thank you Joseph! -Mark $\endgroup$ – Mark Viola Jan 28 '17 at 21:12
  • $\begingroup$ 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. $\endgroup$ – zhw. Jan 28 '17 at 21:33
  • $\begingroup$ @zhw. Yes, I suppose it is since $n^2$ is an integer. I've deleted the preamble. $\endgroup$ – Mark Viola Jan 28 '17 at 21:37
  • $\begingroup$ @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.. $\endgroup$ – Abhishek Chandra Jan 29 '17 at 5:36

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