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Question:

Show that the radius of convergence of $\sum_{k=0}^\infty c_kx^k$ is $\lim_{n\to\infty} \frac{c_n}{c_{n+1}}$ if this limit exists

My attempt:

The only thought that occurred to me was to attempt the ratio test:

$\lim_{k\to\infty} |\frac{c_{k+1}x^{k+1}}{c_kx^k}| = \lim_{k\to\infty} |\frac{c_{k+1}\not x^k x}{c_k\not x^k}| = x\lim_{k\to\infty}\frac{c_{k+1}}{c_k}$

But I am unsure of this being the correct method, or how to connect what I am doing with the conclusion I'm seeking.

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  • $\begingroup$ Please add the definition of $s_n$ to your question. It is probably the sum of the first $n$ terms of the sequence $c_k$? $\endgroup$ – Justpassingby Dec 8 '15 at 20:37
  • $\begingroup$ Sorry. I didn't notice the discrepancy. I am looking at the question right now; it's not explicitly defined in any of the questions. However, in previous proofs, $s_n$ represents the sum of the first n terms, so I think you are right. Would it be safe to assume that? $\endgroup$ – lollercide Dec 8 '15 at 20:43
  • $\begingroup$ I would assume it represents the sum of the first $n$ terms without taking $x$ on board. The existence of the limit should not depend on $x.$ $\endgroup$ – Justpassingby Dec 8 '15 at 20:45
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Lecornu, french mathematician, have proved that, if the ratio $c_{n+1}/c_n$ has a limit L then L is a singular point of $f(x)=\sum_{n=0}^\infty c_n x^n$. But this is not necessary the unique singular point on the circle of convergence.

(counterexample: $f(x) = \frac{x}{1-x}+\ln(1+x) =\sum_{m=1}^\infty \left(1+\frac{(-1)^m}{m}\right)x^m$ have two singular points, 1 and -1, but $c_{n+1}/c_n$ converges to 1.)

see Hadamard, sur le rayon de convergence des séries ordonnées selon les puissances d'une variable, Comptes rendus de l'académie des sciences de Paris, 23 janvier 1888.

Hadamard, La série de Taylor et son prolongement analytique, Scientia, 1903.

Hadamard & Mandelbrot, La série de Taylor et son prolongement analytique, 1928.

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  • $\begingroup$ But doesn't this counterexample conform to the radius of convergence, just not a single point of convergence? $\endgroup$ – lollercide Dec 9 '15 at 13:15
  • $\begingroup$ if you know the singular points, you know the radius of convergence: this is the smallest distance between 0 and the singular points. $\endgroup$ – Claudeh5 Dec 9 '15 at 16:40

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