# Prove the limit exists

I was toying around with Abel summability when I stumbled upon a limit I could not prove existed.

$$\lim_{x\to-1^+}\sum_{n=2}^\infty x^n\ln(n)\tag{*}$$

While it may not be clear that such a limit could converge, it may be helpful to note a similar example:

$$\lim_{x\to-1^+}\sum_{n=1}^\infty nx^{n-1}=\lim_{x\to-1^+}\frac1{(1-x)^2}=\frac14$$

However, a lack of closed form of $(*)$ makes it difficult for me to show it converges. WolframAlpha returns the series as a derivative of the Lerchphi function, though it doesn't seem quite helpful.

By considering

$$f_k(x)=\sum_{n=2}^k x^n\ln(n)$$

I find that

$$f_{35}(-0.75)-0.25f'_{35}(-0.75)=0.225803586648$$

Which is a quick linear approximation of $f_{35}(x)$ centered at $x=-\frac34$. This agrees with what I think to be the limit:

$$\lim_{x\to-1^+}\sum_{n=2}^\infty x^n\ln(n)\stackrel?=\eta'(0)=\frac12\ln\left(\frac\pi2\right)=0.225791352645\tag{**}$$

Where $\eta(s)$ is the Dirichlet eta function.

I also tried considering more elementary approaches to showing the limit exists, such as using $\ln(n+1)=\ln(n)+\mathcal O(n^{-1})$, however, I could not make use of it.

Bonus points if you can prove $(**)$.

• The sum for $f_n(x)$ doesn't make sense because $n$ is both the index of summation and the upper limit. – marty cohen Jun 21 '17 at 1:27
• @martycohen Oops, is an obvious typo on my part. – Simply Beautiful Art Jun 21 '17 at 1:29
• language comment: "could converge" should be "could exist and be finite" But very nice write up! – zhw. Jun 21 '17 at 1:29
• @zhw. Hm, what is the difference between a limit converging and a limit existing and being finite? And thank you! – Simply Beautiful Art Jun 21 '17 at 1:30
• We wouldn't say the limit of $(\sin x)/x$ converges to $1$ as $x\to 0.$ We might say $(\sin x)/x$ converges to $1$ as $x\to 0.$ Or we might say the limit of $(\sin x)/x$ as $x\to 0$ is $1.$ The point is: the limit is not moving around, doing pirouettes and cartwheels, and then finally settling down to a steady state. The limit either exists or it doesn't. If it exists, it's a number. – zhw. Jun 21 '17 at 1:41

Recall the archetypal Frullani integral

$$\int_{0}^{\infty} \frac{e^{-u} - e^{-nu}}{u} \, du = \log n.$$

Since the integrand is non-negative for all $n \geq 1$, when $x \in [0, 1)$ we can apply the Tonelli's theorem to interchange the sum and integral unconditionally to get

\begin{align*} \sum_{n=1}^{\infty} x^n \log n &= \int_{0}^{\infty} \left( \sum_{n=1}^{\infty} x^n \cdot \frac{e^{-u} - e^{-nu}}{u} \right) \, du \\ &= \frac{x^2}{1-x} \int_{0}^{\infty} \frac{e^{-u}(1 - e^{-u})}{u(1-xe^{-u})} \, du. \tag{1} \end{align*}

Notice that the last integral converges absolutely. So this computation can be fed back to the Fubini's theorem, showing that exactly the same computation can be carried out to prove $\text{(1)}$ for all $|x| < 1$.

Now we would like to take limit as $x \to -1^{+}$. When $x \in (-1, 0]$, the integrand of the last integral of $\text{(1)}$ is uniformly bounded by the integrable function $u^{-1}e^{-u}(1-e^{-u})$. Therefore by the dominated convergence theorem, as $x \to -1^{+}$ we have

$$\lim_{x\to -1^+} \sum_{n=1}^{\infty} x^n \log n = \frac{1}{2} \int_{0}^{\infty} \frac{e^{-u}(1 - e^{-u})}{u(1+e^{-u})} \, du.$$

This already proves that the limit exists, but it even tells more that the limit is indeed $\eta'(0)$. To this end, we first perform integration by parts to remove the pesky factor $u$ in the denominator. Then the right-hand side becomes

$$\frac{1}{2} \int_{0}^{\infty} \frac{e^{-u}(1 - e^{-u})}{u(1+e^{-u})} \, du = \int_{0}^{\infty} \left( \frac{e^u}{(e^u + 1)^2} - \frac{e^{-u}}{2} \right) \log u \, du$$

In order to compute this integral, it suffices to prove the following claim.

Claim. We have

$$\int_{0}^{\infty} e^{-u} \log u = -\gamma, \qquad \int_{0}^{\infty} \frac{e^u \log u}{(e^u + 1)^2} \, du = -\frac{\gamma}{2} + \eta'(0).$$

Notice that the first claim is an immediate consequence of the identity $\psi(1) = -\gamma$, where $\psi$ is the digamma function. Next, term-wise integration gives

$$\int_{0}^{\infty} \frac{u^{s-1}}{e^{\alpha u} + 1} \, du = \alpha^{-s} \Gamma(s)\eta(s)$$

where $\alpha > 0$ and $s$ is initially assumed to satisfy $\Re(s) > 1$ (so that interchanging the summation and integration works smoothly). Differentiating both sides w.r.t. $\alpha$ gives

$$\int_{0}^{\infty} \frac{u^s e^{\alpha u}}{(e^{\alpha u} + 1)^2} \, du = \alpha^{-s-1} \Gamma(s+1)\eta(s).$$

Although we initially assumed $\Re(s) > 1$, now both sides define an analytic function for $\Re(s) > -1$, hence by the principle of analytic continuation this identity extends to this region as well. Now plugging $\alpha = 1$ and differentiating both sides w.r.t. $s$, we get

$$\int_{0}^{\infty} \frac{u^s e^u \log u}{(e^u + 1)^2} \, du = \Gamma(s+1)\psi(s+1)\eta(s) + \Gamma(s+1)\eta'(s).$$

Plugging $s = 0$ and using known values $\psi(1) = -\gamma$ and $\eta(0) = \frac{1}{2}$, this yields

$$\int_{0}^{\infty} \frac{e^u \log u}{(e^u + 1)^2} \, du = -\frac{\gamma}{2} + \eta'(0).$$

• @SimplyBeautifulArt, Thank you. Now I fixed my computation. – Sangchul Lee Jun 21 '17 at 2:23
• It looks like he used Borel Summability from the way he interchanged the summation and the integral, there are other techniques similar to Borel Summability that would work in the case of this problem. – Zophikel Jun 21 '17 at 2:39
• @Zophikel, I admit that I deliberately ignored the issue of justification. Now I supplied the argument for limit computation. – Sangchul Lee Jun 21 '17 at 3:01
• (+1) This is exactly the way I would have followed. Great answer. – Jack D'Aurizio Jun 21 '17 at 14:09
• @SangchulLee with Borel Summation there's different ways to apply the technique in fact a weaker from of a proof using a weaker version of Borel Summation. – Zophikel Jun 21 '17 at 15:21

$\lim_{x\to-1^+}\sum_{n=2}^\infty x^n\ln(n)\tag{$*$}$

With Wolfy's help, I confirmed that the limit is $\frac12\ln(\pi/2) \approx 0.225791...$.

If we just plug in $x = -1$, this is

$\begin{array}\\ \sum_{n=2}^{2m+1} (-1)^n\ln(n) &=\sum_{n=1}^m (\ln(2n)-\ln(2n+1))\\ &=\sum_{n=1}^m \ln(\frac{2n}{2n+1})\\ &=\sum_{n=1}^m -\ln(\frac{2n+1}{2n})\\ &=-\sum_{n=1}^m \ln(1+\frac{1}{2n})\\ &=-\sum_{n=1}^m (\frac{1}{2n}+O(\frac{1}{n^2}))\\ &=-\frac12 \ln(m) + O(1)\\ \end{array}$

The sum to $2m+2$ would then be

$\begin{array}\\ -\frac12 \ln(m) + O(1)+\ln(2m+2) &=-\frac12 \ln(m)+\ln(2)+\ln(m+1) + O(1)\\ &=-\frac12 \ln(m)+\ln(2)+\ln(m)+\ln(1+1/m) + O(1)\\ &=\frac12 \ln(m)+O(1)\\ \end{array}$

Therefore the sum of the even and odd sums is $O(1)$; with a little more work I could get a more precise estimate (by expanding $\ln(1+\frac{1}{2n})$ it looks like the sum would involve $\gamma$ and $\sum (-1)^k\zeta(k)/k$ ).

So, by my sloppy thinking, the Cesaro sum converges so the limit exists.

Here is a more accurate version of the computation.

Lets look at the partial sums.

$\begin{array}\\ \sum_{n=2}^{2m+1} (-1)^n\ln(n) &=\sum_{n=1}^m (\ln(2n)-\ln(2n+1))\\ &=\sum_{n=1}^m \ln(\frac{2n}{2n+1})\\ &=\sum_{n=1}^m -\ln(\frac{2n+1}{2n})\\ &=-\sum_{n=1}^m \ln(1+\frac{1}{2n})\\ &=-\sum_{n=1}^m \sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{k(2n)^k})\\ &=-\sum_{k=1}^{\infty}\frac{(-1)^{k=1}}{k2^k}\sum_{n=1}^m \frac{1}{n^k})\\ &=-\sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{k2^k}\sum_{n=1}^m \frac{1}{n^k})\\ &=-\frac12\sum_{n=1}^m \frac{1}{n}-\sum_{k=2}^{\infty}\frac{(-1)^{k-1}}{k2^k}\sum_{n=1}^m \frac{1}{n^k})\\ &\to -\frac12(\ln(m)+\gamma+o(1))+\sum_{k=2}^{\infty}\frac{(-1)^{k}\zeta(k)}{k2^k}\\ &\to -\frac12(\ln(m)+\gamma)+C+o(1)\\ &\to -\frac12(\ln(m)+\gamma)+\frac12\gamma+ \frac12\ln(\pi) - \ln(2) +o(1)\\ &= -\frac12\ln(m)+ \frac12\ln(\pi/4) +o(1)\\ \end{array}$

since, according to Wolfy, $C =\sum_{k=2}^{\infty}\frac{(-1)^{k}\zeta(k)}{k2^k} = \frac12\gamma+ \frac12\ln(\pi) - \ln(2) \approx 0.167825$.

The sum to $2m+2$ would then be

$\begin{array}\\ -\frac12\ln(m)+ \frac12\ln(\pi/4)+ o(1)+\ln(2m+2) &=-\frac12\ln(m)+ \frac12\ln(\pi/4)+ o(1)+\ln(2)+\ln(m+1)\\ &=-\frac12 \ln(m)+ \frac12\ln(\pi/4)+ o(1)+\ln(2)+\ln(m)+\ln(1+1/m)\\ &=\frac12 \ln(m) + \frac12\ln(\pi)+o(1)\\ \end{array}$

Therefore the sum of the even and odd sums is $\ln(\pi)-\ln(2)+o(1)$. Therefore the average of the first $m$ terms goes to $\frac12\ln(\pi/2) \approx 0.225791...$

So, the Cesaro sum converges so the limit exists and the limit is $\frac12\ln(\pi/2) \approx 0.225791...$

• I did actually consider Cesaro summing, as I know it is weaker and ensures the Abel sum converging, but didn't make much headway. Now that I look at your answer, I feel ashamed for making it to the logarithms and not expanding a little more. – Simply Beautiful Art Jun 21 '17 at 1:55
• For other readers, it may be helpful to point out that: $$\lim_{x\to-1^+}\sum_{n=2}^\infty x^n\ln(n)=\lim_{N\to\infty}\frac1N\sum_{k=2}^N\sum_{n=2}^k(-1)^n\ln(n)$$Assuming the RHS converges. "If we just plug in $x=-1$..." is a bit confusing. – Simply Beautiful Art Jun 21 '17 at 1:56
• Interesting conjecture, this actually holds true see here for details: math.stackexchange.com/questions/2058805/… – Zophikel Jun 21 '17 at 2:50

This post is to address partial proof of the conjecture, stated by @SimplyBeautifulArt in the comments section using a weaker approach via the Borel Summability, and Euler Summation. The initial attack on the problem starts from $\text{Lemma} \, (1.0)$

$\text{Lemma} \, (1.0)$

$\text{Euler Summation}$

One considers the Divigernt Series $\sum_{}a_{n}$we replace our divigrent series with the corresponding power series:$\sum_{} a_{n}x^{n}$ If such series is convergent for $|x| < 1$ and if it's limit $x \rightarrow 1^{-}$ then one defines the Euler summation of the orginal series as: $$\text{E}( \sum_{n} a_{n}) = \lim_{x \rightarrow 1^{-}} a_{n}x^{n}.$$

$\text{Proposition} \, (1.1)$

$$\lim_{x\to-1^+}\sum_{n=2}^\infty x^n\ln(n)=\lim_{N\to\infty}\frac1N\sum_{k=2}^N\sum_{n=2}^k(-1)^n\ln(n).$$

$\text{Remark}$:

We assume the RHS side converges in $\text{Proposition} \, \, (1.1).$

$\text{Lemma} \, \, (1.1)$

One can observe in our original Proposition the corresponding series on the RHS side can be rewritten as follows

$(1)$

$$\lim_{x\to-1^+}\sum_{n=2}^\infty x^n\ln(n)= \lim_{N \rightarrow \infty} \frac{1}{N} \sum_{k}(-1)^{n} + \sum_{n} \ln(n).$$

Now focusing our observations on the RHS side of our recent result, another key consideration that can be made is by applying Borel Summability as follows

$(2.)$

$$\lim_{x\to-1^+}\sum_{n=2}^\infty x^n\ln(n)= \lim_{n \rightarrow \infty} \frac{1}{N}(\lim_{t \rightarrow \infty} e^{-t} \sum_{k}\frac{t^{n}}{n!}(\sum_{n}(-1)^{n}) + \lim_{x \rightarrow 1^{-}}\sum_{n} \ln(x)^{n})$$

$\text{Remark}$

The recent development seen in $(2.)$ can't just be achieved with Borel Summability alone the series $\sum\ln(n)$ was dealt with via Euler Summation as formally discussed in $(0.0)$. So considering the formalities of Euler Summation the series $\sum\ln(n)$ can be defined as follows in $(3)$

$(3)$

$$\text{E}( \sum_{n} \ln(n)) = \lim_{x \rightarrow 1^{-}} \sum_{n} \ln(x)^{n}.$$

• Oh and also this paper introduced some interesting methods used to achieve some of the partial results:arxiv.org/pdf/1703.05164.pdf – Zophikel Jun 21 '17 at 16:35
• $\lim_{x\to-1^+}\sum_{n=2}^\infty x^n\ln(n)= \lim_{n \rightarrow \infty} \frac{1}{N}(\lim_{t \rightarrow \infty} e^{-t} \sum_{k}\frac{t^{n}}{n!}(\sum_{n}(-1)^{n}) + \lim_{x \rightarrow 1^{-}}\sum_{n} \ln(x)^{n})$, when dealing with this key result, since I forgot to address this in the post it's possible to write the sums:$\sum_{k}\frac{t^{n}}{n!}(\sum_{n}(-1)^{n}) + \lim_{x \rightarrow 1^{-}}\sum_{n} \ln(x)^{n})$ as a single series as opposed to individual series' so taking this key development into consideration we have the following observation. – Zophikel Jun 21 '17 at 17:04
• $$\lim_{x\to-1^+}\sum_{n=2}^\infty x^n\ln(n)= \lim_{t \rightarrow \infty} \frac{1}{N}(e^{-t}\sum_{k}\frac{t^{n}}{n!} \lim_{x \rightarrow 1-}\sum_{n} \sum_{n}(-1)^{n} \ln(x)^{n}$$ – Zophikel Jun 21 '17 at 17:05
• When dealing with a result that's in this form it's possible to use Tauberian technique's when dealing with multiple series i.e(Double, Triple, etc), I encourage users to explore this route with the groundwork I've laid out. – Zophikel Jun 21 '17 at 17:24

For $|z| \le 1, \Re(s)> 1$ and for $|z| < 1$ let the polylogarithm $$Li_s(z) = \sum_{n=1}^\infty n^{-s}z^n$$ For $\Re(s) > 1$ we have $Li_s(1) = \zeta(s), Li_s(-1) = -\eta(s)$.

Now for $|z| \le 1, z \ne 1$, summation by parts shows that $Li_s(z)$ is entire in $s$ and continuous in $z$ and hence $$-\eta(s) = \lim_{z \to -1, |z| \le 1} Li_s(z), \qquad -\eta'(s) = \lim_{z \to -1, |z| \le 1} \frac{\partial}{\partial s}Li_s(z)$$ $$\lim_{z \to -1, |z| \le 1} \sum_{n=1}^\infty z^n \log n = -\eta'(0) = \frac{\log(\pi/2)}{2}$$ (see there)