Limit of a power series in $\beta$ multiplied by $(1 - \beta)$

Question

Suppose that you are given a bounded sequence of real numbers $|w_k| \le W$.

What should be the limit $\lim_{\beta \rightarrow 1^-}\ (1 - \beta) \sum_{k = 0}^\infty \beta^k w_k$?

To see that the limit exists, consider that the function $v(\beta) = (1 - \beta) \sum_{k = 0}^\infty \beta^k w_k$ is analytic and that $|v(\beta)| \le (1 - \beta) \sum_{k = 0}^\infty W \beta^k \le W$ so it is bounded near $\beta = 1$ and admits a limit.

I guess that is should be something like $\limsup_n \frac1n \sum_{k = 0}^{n - 1} w_k$ but I failed to prove it.

Edit: The limit could not exist, see the answer of metamorphy. I'm still interested if the relation $$ \limsup_{\beta \rightarrow 1^-} \ (1 - \beta) \sum_{k = 0}^\infty \beta^k w_k = \limsup_n \frac1n \sum_{k = 0}^{n - 1} w_k$$ holds or not.

Answer 1

Regarding your guess of the limit, if everything converges nicely (as power series with bounded coefficients do for $|\beta|< 1$) then

$$ (1-\beta)\sum_{k=0}^\infty w_k\beta^k = \sum_{k=0}^\infty (w_k-w_{k-1})\beta^k, \quad (|\beta| < 1), $$ after the mild abuse of notation $w_{-1}:=0$. Thus if the limit does exist, it has to be the Abel sum of the backward differences $\Delta w_k:= w_{k}-w_{k-1}$. c.f. definition of Abel summation on Wikipedia.

In particular, if I understand it correctly, metamorphy's example is a variant of the usual example of how some generalised summation methods are affected by dilution of the summands(i.e. padding the sequence $\Delta w_k$ with a bunch of zeros so that their sum $w_k$ stays constant for extended stretches of $k$). See for instance this paper I found on a google search and of course in the book by Hardy, "Divergent Series" (page 59). Also, the divergence result of metamorphy is proven on page 93 with slightly more generality (I'll quote the proof at the end.)

In any case, your guess is like saying that the limit is equal to the Cesàro sum of $\Delta w_k$. Although Cesàro and Abel sums agree when defined, there are times when Cesàro doesn't exist but Abel does. But from Theorem 92 of Divergent Series, your guess is true (with lim and not just limsup) the moment the LHS exists, since $\Delta w_k$ are bounded. That is, we have

Suppose $\Delta w_k $ is bounded. Then these two limits exist are equal as long as either one exists: $$ \lim_{\beta \rightarrow 1^{-}}(1-\beta) \sum_{k=0}^{\infty} \beta^{k} w_{k}=\lim _{n\to\infty} \frac{1}{n} \sum_{k=0}^{n-1} w_{k}$$

(Boundedness is essential. For instance, $\Delta w_k = (-1)^k(k+1)$ has Abel sum $1/4$ but is not Cesàro summable, and also the limsup isn't equal to the Abel sum, since the average of the partial summands alternate between $ k/(2k-1) \approx 1/2$ and $0$. )

For example, the "undiluted" version of metamorphy's example (up to a change in $\Delta w_0$ and a scaling) is the Cesàro summable $ \Delta w_k $ given by $$ \Delta w_k = (-1)^{k}, \quad w_k = \frac{(-1)^k+1}2,$$

and the RHS of the claimed identity is $\lim _{n\to\infty} \frac{1}{n} \sum_{k=0}^{n-1} w_{k} = 1/2$. The LHS is $$\lim_{\beta\uparrow 1}(1-\beta)\sum_{k=0}^\infty \beta^{2k} = \lim_{\beta\uparrow 1}\frac{1-\beta}{1-\beta^2}= \lim_{\beta\uparrow 1} \frac{1}{1+\beta} =\frac12,$$ as predicted.

Finally, I wanted to quote wholesale the beautiful proof of Hardy from page 93; I think more people should read it. It proves that Abel summability can be destroyed by dilution of series (which is a stronger result than the analogous result for Cesàro summability.)

On the other hand, if $a>1,$ then $$ F(x)=x-x^{a}+x^{a^{2}}-x^{a^{2}}+\dots$$ does not tend to a limit when $x \rightarrow 1$ . To see this, we observe that $F(x)$ satisfies the functional equation $$F(x)+F\left(x^{a}\right)=x$$ and that $$ \Phi(x)=\sum \frac{(-1)^{n}}{n !\left(1+a^{n}\right)}\left(\log \frac{1}{x}\right)^{n}$$ is another solution. Hence $\Psi(x)=F(x)-\Phi(x)$ satisfies $\Psi(x)=-\Psi\left(x^{a}\right),$ and is therefore a periodic function of $\log \log (1 / x)$ with period $2 \log a .$ since it is plainly not constant, it oscillates between finite limits of indetermination when $x \rightarrow 1,\log (1 / x) \rightarrow 0, \log \log (1 / x) \rightarrow-\infty .$ But $\Phi(x) \rightarrow \frac{1}{2},$ and therefore $F(x)$ oscillates.

It follows that $1-1+1-\ldots$ is not summable $(A, \lambda)$ when $\lambda_{n}=a^{n}(a>1)$.

Answer 2

Of course you can't prove it already because of expected behaviour of the answer w.r.t. $w_k\mapsto-w_k$. Further, your "... and admits a limit" is left unjustified, and in fact the limit can fail to exist.

Here is a counterexample, maybe not the simplest but I think interesting: $$w_k=\begin{cases}0,&k=0\\\color{blue}{(-1)^{\lfloor\log_2 k\rfloor}},&k>0\end{cases}$$ with $w(\beta):=(1-\beta)\sum_{k=0}^{\infty}w_k\beta^k=\beta+2\sum_{n=1}^{\infty}(-1)^n\beta^{2^n}$. To analyse $\beta\to 1^{-}$, let's consider $W(x)=w(e^{-x})$ for $x>0$, and use $$e^{-y}=\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}\Gamma(s)y^{-s}\,ds$$ where $y,c>0$ are arbitrary. We get $$W(x)=\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}\frac{\Gamma(s)}{x^s}\frac{2^s-1}{2^s+1}\,ds=A(x)+B(x),$$ where $A(x)$ and $B(x)$ come from residues at $s=-n$ and $s=\pm(2n+1)\pi i/\ln 2$: $$A(x)=\sum_{n=1}^{\infty}(-1)^{n-1}\frac{2^n-1}{2^n+1}\frac{x^n}{n!},\\ B(x)=\frac{4}{\ln 2}\sum_{n=0}^{\infty}\Re\left[e^{-(2n+1)\pi i\log_2 x}\Gamma\Big(\frac{2n+1}{\ln 2}\pi i\Big)\right].$$ While $A(x)\to 0$ with $x\to 0$, $B(x)$ keeps oscillating, with no convergence to anything.

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Regarding the "Edit": nope. The same counterexample works. Namely, $$|B(x)|\leqslant\frac{4}{\ln 2}\sum_{n=0}^{\infty}\left|\Gamma\Big(\frac{2n+1}{\ln 2}\pi i\Big)\right|<0.0055$$ while $\displaystyle\limsup_{n\to\infty}\frac{1}{n}\sum_{k=0}^{n-1}w_k=\frac{1}{3}$ (considering the subsequence with $n=2^{2k+1}$).