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I am interested in better understanding "extensions" of power series beyond their radius of convergence.

For example, $$\log(1+x) = \sum_{k=0}^\infty \frac{(-1)^k}{k+1} x^{k+1}$$ only holds for $x$ within the unit disk, while $$\log(1+x) = \lim_{n \to \infty} \sum_{k=0}^n \frac{(-1)^k}{k+1} x^{k+1} R(k,n)$$ for an appropriate regularizer $R(k,n)$ can hold for a different set of $x$.

I am curious about the regularizer $$R(k,n) = B(k,n) := \frac{\binom{2n}{n+k}}{\binom{2n}{n}},$$ which is rapidly decaying as $k$ approaches $n$, and is interpreted as $0$ for $k>n$. With some numerical experiments, it appears to perform quite well.

How can I prove that $$\log(1+x) = \lim_{n \to \infty} \sum_{k=0}^n \frac{(-1)^k}{k+1} x^{k+1} B(k,n)$$ for all real $x>-1$?

Please feel free to direct me to Hardy's Divergent Series, although please let me know which theorems I should brush up on from there (it has been a long time since I read it). I want to note that the regularizer above is regular, in the sense that the limit sends already-convergent series to their original values. Thus, we are guaranteed that $\log(1+x) = \lim_{n \to \infty} \sum_{k=0}^n \frac{(-1)^k}{k+1} x^{k+1} B(k,n)$ in the interval $-1<x<1$ where the original power series converged. Thus, I believe only real $x>1$ needs to be considered.

Thành Nguyễn
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user196574
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  • I don't know what a regularizer is, but the series of $\ln(\frac{1+x}{1-x})$ can be used to find the natural log of any number by solving for $y = \frac{1+x}{1-x}$. – Supernerd411 May 16 '25 at 05:16

1 Answers1

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For your specific limit, one has $$ \binom{2n}{n}^{-1}\sum_{k=0}^n\binom{2n}{n+k}(-x)^k=\int_0^1(1-x+xt^{1/n})^n\,dt. $$

BTW, this is a particular case of the identity (a variant of the one here) $$\sum_{k=0}^m\binom{n}{k}x^k=(n-m)\binom{n}{m}\int_0^1 t^{n-m-1}(1+x-t)^m\,dt\qquad(0\leqslant m<n)$$ (at $n\gets 2n$, $m\gets n$ and $x\gets-1/x$) after the substitution $t\gets t^{1/n}$.

Now we can integrate this w.r.t. $x$, then take $n\to\infty$ (or vice versa, with some justification).

metamorphy
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  • As a note to myself, I think this result also answers the following question: for which $f(n)$ does $\binom{2n}{n}^{-1}\sum_{k=0}^n\binom{2n}{n+k}(-f(n))^{k+1}/(k+1) \sim \log(f(n))$ at large $n$? For this asymptotics to work out, we need $e^{n \log(1-f(n)+f(n)t^{1/n})} \sim t^{f(n)}$ for fixed $t>0$, and that should happen when $n (f(n)/n)^2 \to 0$; i.e. for $f(n) = o(n^{1/2})$. In particular, it behaves well for $f(n) \sim x^{1/3}$. Thanks again! – user196574 May 16 '25 at 23:03