Evaluating the limit $\lim_{n \to \infty} \left[\log(n) + 2n \log(4n + 2) - \sum_{k = 1}^{2n} (-1)^k (2k+1)\log(2k+1)\right]$

Question

I am trying to show that the limit $$\lim_{n \to \infty} \left[\log(n) + 2n \log(4n+2) - \sum_{k=1}^{2n} (-1)^k (2k+1) \log(2k + 1)\right] = \frac{2G}{\pi} - \log(4)$$ where $G$ is Catalan's constant.

My attempt was to turn the partial sum into a product: \begin{align*} \sum_{k=1}^{2n} (-1)^n (2k+1) \log(2k + 1) &= \log\left(\prod_{k=1}^{2n}(2k+1)^{(-1)^k(2k+1)} \right). \end{align*} \begin{align*} a_n := \prod_{k=1}^{2n}(2k+1)^{(-1)^k(2k+1)} \end{align*} \begin{align*} a_n &= \frac{\left(\prod_{k=1}^{n} (4k+1)^{(4k+1)}\right)^2}{\prod_{k=1}^{2n} (2k+1)^{(2k+1)}} \\ &=\frac{\left(\prod_{k=1}^{n} (4k+1)^{(4k+1)}\right)^2}{\prod_{k=1}^{2n} (2k+1)^{(2k+1)}} \cdot \frac{4^{n (2 n+1)} (H(2 n))^2}{\prod_{k = 1}^{2n} (2n)^{2n}} \quad \quad \Big[H(n) = \text{the hyperfactorial} \Big]\\ &= \frac{4^{n (2 n+1)} (H(2 n))^2}{H(4n+1)}\left(\prod_{k=1}^{n} (4k+1)^{(4k+1)}\right)^2. \end{align*}

Another attempt, I considered using the following definition of the gamma function $$\Gamma(x + 1) = \lim_{n \to \infty} \frac{n^x}{\prod_{k = 1}^n \left(1 + \frac{x}{k} \right)},$$ but I was unable to fund any helpful algebraic moves. I also considered using Abel's summation formula to simplify the partial sum down to a closed form, but quickly realized it was not helpful. From here, I was unable to find a closed form.

I also found this paper that I believe may be relevant to this problem. A note on products involving ζ(3) and Catalan’s constant. Are there any algebraic moves that can be done using the gamma function's infinite product representation to derive the result? Is it possible there is an asymptotic formula that can be substituted in to evaluate this limit? Is there a better way to evaluate this limit? If so, what can be done?

Answer 1

I derive an asymptotic expansion with explicit error bound for the sum in question. Employing the Euler–Boole summation formula with $f(x) = 2x\log (2x)$, $h = \frac{1}{2}$, $n=2n+1$, $a=1$ and $m=2$, we obtain \begin{align*} & \sum\limits_{k = 1}^{2n} {( - 1)^k (2k + 1)\log (2k + 1)} = \frac{1}{2}(4n + 2)\log (4n + 2) - \log 2 + \int_1^{2n + 1} {\frac{1}{x}\widetilde E_1 \!\left( {\tfrac{1}{2} - x} \right){\rm d}x} \\ & = \frac{1}{2}(4n + 2)\log (4n + 2) - \log 2 + \int_1^{ + \infty } {\frac{1}{x}\widetilde E_1 \!\left( {\tfrac{1}{2} - x} \right){\rm d}x} - \int_{2n + 1}^{ + \infty } {\frac{1}{x}\widetilde E_1 \!\left( {\tfrac{1}{2} - x} \right){\rm d}x} . \end{align*} The improper integrals are convergent due to the Dirichlet criteria, and thus the second integral tends to $0$ as $n\to +\infty$. The value of the first integral may be obtained via Fourier expansion: \begin{align*} \int_1^{ + \infty } {\frac{1}{x}\widetilde E_1 \!\left( {\tfrac{1}{2} - x} \right){\rm d}x} & = \log 2 + \int_0^{ + \infty } {\frac{1}{x}\widetilde E_1 \!\left( {\tfrac{1}{2} - x} \right){\rm d}x} \\ & = \log 2 - \frac{4}{{\pi ^2 }}\sum\limits_{k = 0}^\infty {\frac{{( - 1)^k }}{{(2k + 1)^2 }}\int_0^{ + \infty } {\frac{{\sin ((2k + 1)\pi x)}}{x}{\rm d}x} } \\ & = \log 2 - \frac{2}{\pi }\sum\limits_{k = 0}^\infty {\frac{{( - 1)^k }}{{(2k + 1)^2 }}} = \log 2 - \frac{{2G}}{\pi }, \end{align*} where $G$ is Catalan's constant. Accordingly, $$ \sum\limits_{k = 1}^{2n} {( - 1)^k (2k + 1)\log (2k + 1)} = \frac{1}{2}(4n + 2)\log (4n + 2) - \frac{{2G}}{\pi } + o(1) $$ as $n\to +\infty$. If we use the Euler–Boole summation formula with the same input but with $m=2m+1$ and note that odd Euler polynomials vanish at $h=\frac{1}{2}$, we find \begin{align*} \sum\limits_{k = 1}^{2n} {( - 1)^k (2k + 1)\log (2k + 1)} & = \frac{1}{2}(4n + 2)\log (4n + 2) + C_m + \frac{1}{2}\sum\limits_{k = 1}^{m} {\frac{{E_{2k} }}{{(2k)!2^{2k} }}f^{(2k)} (2n + 1)} \\ & \quad\, - \frac{1}{{2(2m)!}}\int_{2n + 1}^{ + \infty } {f^{(2m+1)} (x)\widetilde E_{2m} \!\left( {\tfrac{1}{2} - x} \right){\rm d}x} \\ & =\frac{1}{2}(4n + 2)\log (4n + 2) + C_m + \sum\limits_{k = 1}^{m} {\frac{{E_{2k} }}{{4k(2k - 1)}}\frac{1}{{(4n + 2)^{2k - 1} }}} \\ & \quad\, + \frac{1}{{2m}}\int_{2n + 1}^{ + \infty } {\frac{1}{{x^{2m} }}\widetilde E_{2m}\! \left( {\tfrac{1}{2} - x} \right){\rm d}x} \end{align*} with some constant $C_m$ coming from the terms of the summation formula that do not depend on $n$. The last integral clearly converges to $0$ as $n\to +\infty$, hence, by our previous result, we must have $C_m =- \frac{{2G}}{\pi }$ for all $m$. We now note that $$ \frac{1}{{2m}}\int_{2n + 1}^{ + \infty } {\frac{1}{{x^{2m} }}\widetilde E_{2m}\! \left( {\tfrac{1}{2}- x} \right){\rm d}x} = - \frac{1}{{2m}}\int_0^{ + \infty } {\frac{1}{{(t + 2n + 1)^{2m} }}\widetilde E_{2m}\! \left( {\tfrac{1}{2}- t} \right){\rm d}t} $$ and $$ \frac{{E_{2m} }}{{4m(2m - 1)}}\frac{1}{{(4n + 2)^{2m - 1} }} = \frac{1}{{2m}}\int_0^{ + \infty } {\frac{{2^{ - 2m} E_{2m} }}{{(t + 2n + 1)^{2m} }}{\rm d}t}, $$ i.e., \begin{align*} \sum\limits_{k = 1}^{2n} {( - 1)^k (2k + 1)\log (2k + 1)} = \frac{1}{2}(4n + 2)\log (4n + 2) - \frac{{2G}}{\pi } & + \sum\limits_{k = 1}^{m-1} {\frac{{E_{2k} }}{{4k(2k - 1)}}\frac{1}{{(4n + 2)^{2k - 1} }}} \\ & +R_m(n) \end{align*} with $$ R_m(n)=\frac{1}{{2m}}\int_0^{ + \infty } {\frac{{2^{ - 2m} E_{2m} - \widetilde E_{2m} (1/2 - t)}}{{(t + 2n + 1)^{2m} }}{\rm d}t} . $$ Since $$ \left| {2^{ - 2m} E_{2m} - \widetilde E_{2m} (1/2 - t)} \right| \le 2^{ - 2m} \left| {E_{2m} } \right| + \sup _{0<x<1/2} \left| {E_{2m} (x)} \right| \le 2 \cdot 2^{ - 2m} \left| {E_{2m} } \right|, $$ we find that $$ |R_m (n)| \le 2 \cdot \frac{{\left| {E_{2m} } \right|}}{{4m(2m - 1)}}\frac{1}{{(4n + 2)^{2m - 1} }}. $$ Thus, in summary, \begin{multline*}\boxed{ \sum\limits_{k = 1}^{2n} {( - 1)^k (2k + 1)\log (2k + 1)} = \frac{1}{2}(4n + 2)\log (4n + 2) - \frac{{2G}}{\pi } + \sum\limits_{k = 1}^{m-1} {\frac{{E_{2k} }}{{4k(2k - 1)}}\frac{1}{{(4n + 2)^{2k - 1} }}} +R_m(n)} \end{multline*} for any $n\ge 1$, $m\ge 1$, with $$\boxed{ |R_m (n)| \le 2 \cdot \frac{{\left| {E_{2m} } \right|}}{{4m(2m - 1)}}\frac{1}{{(4n + 2)^{2m - 1} }}.} $$ I expect that with more work, the factor $2$ in the error bound can be removed and it can be shown that $R_m(n)$ has the same sign as the first omitted term.

Answer 2

Switching $n\to2n$ for the convenience, we have to prove that $$\lim_{n\to\infty}\Big(\ln\frac{n}{2}+n\ln(2n+2)-\sum_{k=1}^n(-1)^k(2k+1)\ln(2k+1)\Big)=\frac{2G}{\pi}-2\ln2\quad (n\,\text{is even})$$ or, considering the asymptotics of the sum, to prove that $$S(n)=\sum_{k=1}^n(-1)^k(2k+1)\ln(2k+1)=n\ln n+n\ln2+\ln n+\ln2+1-\frac{2G}{\pi}+o(1)$$ Using the Frullani integral and performing summation, we can present the sum in the form (for even $n$) $$S(n)=\sum_{k=1}^n(-1)^k(2k+1)\int_0^\infty\frac{e^{-t}-e^{-(2k+1)t}}{t}dt$$ $$=\int_0^\infty\left(ne^{-t}+\frac{e^{-t}}{1+e^{2t}}+\frac{2e^t}{(1+e^{2t})^2}-\frac{(2n+1)e^{-(2n+1)t}}{1+e^{2t}}-\frac{2e^{-(2n-1)t}}{(1+e^{2t})^2}\right)\frac{dt}{t}$$ For the convenience of operation with every term separately, we consider a regularized integral (where $\epsilon$ is small but finite): $$S_\epsilon(n)=\int_0^\infty\Big(...\Big)\frac{dt}{t^{1-\epsilon}}=S_0+S_1+S_2+S_3+S_4\tag{0}$$ Then $$S_0=n\int_0^\infty t^{\epsilon-1}e^{-t}dt=n\,\Gamma(\epsilon)\tag{1}$$ $$S_3=-(2n+1)\int_0^\infty t^{\epsilon-1}\frac{e^{-(2n+1)t}}{1+e^{2t}}dt$$ Making the substitution $t=\frac{x}{2n+1}$ and keeping only non-zero terms at $n\to\infty$ $$S_3=-\frac{(2n+1)^{1-\epsilon}}{2}\int_0^\infty\frac{e^{-x}x^{\epsilon-1}}{1+\frac{x}{2n+1}}dx+o(1)$$ $$=-n\Gamma(\epsilon)-\frac{\Gamma(\epsilon)}{2}+n\ln(2n)+\frac{1}{2}\ln(2n)+1+o(1)\tag{2}$$ In the same way we evaluate $S_4$: $$S_4=-\int_0^\infty\frac{2e^{-(2n-1)t}}{(1+e^{2t})^2}t^{\epsilon-1}dt=-\,\frac{\Gamma(\epsilon)}{2}+\frac{1}{2}\ln(2n)+o(1)\tag{3}$$ Next, let's consider $$S_1+S_2-\Gamma(\epsilon)=\int_0^\infty\Big(\frac{e^{-t}}{1+e^{2t}}+\frac{2e^2}{(1+e^{2t})^2}-e^{-t}\Big)t^{\epsilon-1}dt$$ $$=\int_0^\infty\frac{e^t-e^{3t}}{(1+e^{2t})^2}t^{\epsilon-1}dt=-\frac{1}{2}\int_0^\infty\frac{\sinh t}{\cosh^2t}t^{\epsilon-1}dt$$ We see that this espression is regular at $\epsilon\to0$; therefore, we can put $\epsilon=0$ and consider $$S_1+S_2-\Gamma(\epsilon)\to-\,\frac{1}{4}\int_{-\infty}^\infty\frac{\sinh t}{t\cosh^2t}dt=-\frac{2\beta(2)}{\pi}=-\frac{2G}{\pi}\tag{4}$$ Probably, the easiest way of the evaluation is integration in the complex plane, closing the contour in the upper half-plane and evaluating residues at $t=\pi ik+\frac{\pi i}{2}, k=0,1,2...\,$ Another way is integration by part and usage of the analytical continuation and functional equation for Dirichlet beta-function.

Putting (1)-(4) into (0) we see that all diverging at $\epsilon\to 0$ terms cancel, and we are left with $$\boxed{\,\,S(n)=n\ln(2n)+\ln(2n)+1-\frac{2G}{\pi}+o(1)\,\,}$$ The limit follows.