Prove the limit exists

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

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

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

Added later:

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... $

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 $(3)$

$(3)$

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