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Taken from question 4F Paper 1 Cambridge Tripos 2016:

"Let $(a_n)$ be the number of pairs of integers $(x,y)$ such that $x^2+y^2\le{n^2}$. What is the radius of convergence of the series $\sum_{n=0}^\infty a_nz^n$."

These are the steps in my proof:

1) First, note z=0 results (trivially) in a convergent sum, however, note that if $|z|\ge{1}$, then this series diverges as $\lim{a_n}=\infty$, and $(a_n)$ is a positive, strictly increasing sequence.

2) Next, note $(a_n)\approx{\pi{n^2}}$, so there is some integer k such that $a_n\le{\pi{n^k}}$ for all n.

3) Now, by the Ratio Test, $\sum_{n=0}^\infty{\pi{n^k}} z^n$ is convergent iff $|z|<1$, and so by the Comparison Test, $\sum_{n=0}^\infty a_nz^n$ is also convergent if $|z|<1$. As $\sum_{n=0}^\infty a_nz^n$ is divergent for $|z|\ge{1}$, the radius of convergence for the series is 1.

How is my proof, is it correct? In step 1, I know this is true for positive z, but not entirely sure why for negative z as well... Also, would anyone know how I can rigrously assert the statement in step 2 (I assume its true heuristically)?

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1 Answer 1

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Some comments:

  • In step (1), you know $a_n \ge 1$ since the point $(0,0)$ is always inside the circle. So later you can say $a_n z^n \not \to 0$ for $|z|>1$, and so the sum does not converge to anything

  • In step (2) your assertion is wrong for $n=0$ and $n=1$ though it does not really matter. In any case you do not need to introduce $k$, since for example by putting a square round the circle you can say $1 \le a_n \le (2n+1)^2$ and this will later be enough to get you to a radius of convergence of $1$. Actual values of $a_n$ are in OEIS A000328

  • In step (3) you have sketched a proof but the question also says "You may use the comparison test, provided you state it clearly" so you would need to do that and then apply it explicitly

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  • $\begingroup$ Thank you. In your first bullet point, how does one verify this will be true for negative z, e.g. z=-2, as the sign on $z^n$ will keep changing... $\endgroup$ Jul 4, 2018 at 17:23
  • $\begingroup$ Also (apologies!), where do you get $(2n+1)^2$ from... $\endgroup$ Jul 4, 2018 at 17:29
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    $\begingroup$ @Daniele1234 The set is contained in $\{-n,\dots, -1,0,1,\dots, n\}^2.$ The cardinality of the latter set is $(2n+1)^2$ $\endgroup$
    – zhw.
    Jul 4, 2018 at 17:37
  • $\begingroup$ @Daniele1234 For the sum to converge to anything, the individual terms need to converge to $0$ (I have clarified this in my answer). But since $a_n \ge 1$, you will have $|a_n z^n| \ge 1$ when $|z| \ge 1$ and so $a_n z^n \not \to 0$ and so $\sum a_n z^n$ does not converge $\endgroup$
    – Henry
    Jul 4, 2018 at 17:54

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