-3
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The radius of convergence of the power series $$\sum_{n=0}^\infty a_nz^{n^2}$$ where $a_0 = 1$, $a_n=3^{-n}a_{n-1}$ for $n\in \mathbb{N}$, is

(A) $0$, (B) $\sqrt{3}$, (C) $3$, or (D) $\infty$

In this power series I found that the radius of convergence is $\sqrt{3}$. Am I right? Please justify it.

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    Please share your method? – 2017-02-06

2 Answers 2

1

You are right ! But you have not justified your result !

Prove by induction:

$a_n=\frac{1}{3^{\frac{n(n+1)}{2}}}$. Then consider the series $\sum_{n=0}^{\infty}b_n$, with $b_n=a_nz^{n^2}$.

Then calculate for which $z$ we have $\lim_{n \to \infty}|b_n|^{1/n}<1$ , for which $z$ we have $\lim_{n \to \infty}|b_n|^{1/n}>1$

and for which $z$ we have $\lim_{n \to \infty}|b_n|^{1/n}=1$ .

0

enter image description here This is how I have done . I hope it's right

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    You lost absolute values $|*|$ !! – 2017-02-06
  • 1
    Yes but it will matter only if we find the interval of convergence. Here the radius of convergence will be √3 when we replace the < symbol with = . – 2017-02-06