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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:

- How find this $\lim_{n\to\infty}n^2\left(\frac{1^k+2^k+\cdots+n^k}{n^{k+1}}-\frac{1}{k+1}-\frac{1}{2n}\right)$
- Why is $\lim_{x \to 0} {\rm li}(n^x)-{\rm li}(2^x)=\log\left(\frac{\log(n)}{\log(2)}\right)$?
- How to find this limit with following constraints?
- Limit of sequence in which each term is defined by the average of preceding two terms
- Find functions family satisfying $ \lim_{n\to\infty} n \int_0^1 x^n f(x) = f(1)$
- Evaluation of $ \lim_{x\rightarrow \infty}\left\{2x-\left(\sqrt{x^3+x^2+1}+\sqrt{x^3-x^2+1}\right)\right\}$

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

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 $(**)$.

- Proof of Fourier Inverse formula for $L^1$ case
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- Limit of a periodic function
- Proving limit doesn't exist using the $\epsilon$-$\delta$ definition
- Proving $\left( \sum_{n=-\infty}^{\infty} e^{-\pi n^2} \right)^2= 1 + 4 \sum_{n=0}^{\infty} \frac{(-1)^n}{e^{(2n+1)\pi} - 1}$
- Estimating $\int_0^1f$ for an unknown Lipschitz $f$ to within 0.0001
- Help with $\lim_{x\rightarrow +\infty} (x^2 - \sqrt{x^4 - x^2 + 1})$

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 progress(i.e not a full proof) of the conjecture, stated by @SimplyBeautifulArt in the comments secition using a weaker approach via the Borel Route, and Euler Summation.

$$\text{Euler Summation} \, \, (0.0):$$

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:

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

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

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

Remark: We assume the RHS side converges

$$\text{Lemma}(0.1):$$

One can observed on our original Proposition the corresponding series on the RHS side can be rewritten as follows.

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

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 in $(2.)$

$(2.)$

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

The recent development seen in $(2.)$ can’t just be achived 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:

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

@Simply can you continue from here 🙂

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)

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