Gaining further insight into AMM 11832

This post is aimed as a second solution (v2.0) to AMM problem 11832 proposed by Donald Kunth, U.S.A.

My first proof in a previous blog entry: AMM problem 11832 by D. Knuth

(I believe this new line of computation gains more insight into the cog and wheels (working principles) behind the creation of the identity, rather than my initial approach, which was designed for dealing with a specific case).

We begin with the a theorem with generating functions:

The following is an excerpt from K. N. Boyadzheiv's paper:

Theorem: Given a pair of sequence $\displaystyle (a_n)_{n \ge 0}$ and $\displaystyle (b_n)_{n\ge 0}$ that are Binomial Inverses of each other, that is they satisfy the relation:

$\displaystyle a_n = \sum\limits_{k=0}^{n} (-1)^{n-k}\binom{n}{k} b_k$ and $\displaystyle b_n = \sum\limits_{k=0}^{n} \binom{n}{k} a_k$

for each $n \ge 0$, then the corresponding generating functions with central binomial coefficient weight are related as follows:

If $\displaystyle f_{b}(z) = \sum\limits_{n=0}^{\infty} \binom{2n}{n}b_nz^n$ then, $\displaystyle f_a(z) = \sum\limits_{n=0}^{\infty} \binom{2n}{n}a_nz^n = \frac{f_b\left(\frac{z}{1+4z}\right)}{\sqrt{1+4z}}$

Proof:

Lemma: For complex number $\alpha$ and $|z|$ in suitable region of convergence, we have:

$$\displaystyle \sum\limits_{n=0}^{\infty} (-1)^n\binom{\alpha}{n}a_nz^n = (1+z)^{\alpha}\sum\limits_{n=0}^{\infty}(-1)^n\binom{\alpha}{n}\left(\frac{z}{1+z}\right)^n\left(\sum\limits_{k=0}^{n}\binom{n}{k}a_k\right)$$

which is a consequence of Euler's series transformation formula.

The special case $\displaystyle \alpha = -\frac{1}{2}$ is what we are after, $\displaystyle \binom{-1/2}{n} = \frac{1}{4^n}\binom{2n}{n}$ and making the substitution $z \mapsto 4z$ we get the desired result. [Q.E.D.]

For further details please refer to the original paper itself.

Back to the problem at hand:

We have the Binomial transformation formula: $\displaystyle \sum\limits_{k=1}^{n}(-1)^{n-k}\binom{n}{k}(H_{2k} - H_k) = (-1)^{n-1}\left(\frac{4^n}{2n\binom{2n}{n}} - \frac{1}{2n}\right)$ which can be easily established. Hence, putting it into the machinery provided by the theorem, we get:

$$g(z) := \sum\limits_{n=1}^{\infty} \binom{2n}{n}\left(\frac{4^n}{2n\binom{2n}{n}} - \frac{1}{2n}\right)z^n = \frac{1}{2}\sum\limits_{n=1}^{\infty} \frac{4^nz^n}{n} - \frac{1}{2} \sum\limits_{n=1}^{\infty} \binom{2n}{n}\frac{z^n}{n}= -\frac{1}{2}\log (1-4z) + \log \left(\frac{1+\sqrt{1-4z}}{2}\right)$$

and as a consequence:

$$f(z) := \sum\limits_{n=1}^{\infty} (-1)^{n-1}\binom{2n}{n} (H_{2n} - H_n)z^n = \frac{g\left(\frac{z}{1+4z}\right)}{\sqrt{1+4z}} = \frac{1}{\sqrt{1+4z}}\log \left(\frac{1+\sqrt{1+4z}}{2}\right)$$

Making the change $z \mapsto -z$ in the above identity we get:

$\displaystyle \sum\limits_{n=1}^{\infty} \binom{2n}{n} (H_{2n} - H_n)z^n = \frac{-1}{\sqrt{1-4z}}\log \left(\frac{1+\sqrt{1-4z}}{2}\right)$

Dividing both sides with $z$ and integration leads to D. Knuth's identity, as we saw in my previous proof:

$$\boxed{\sum\limits_{n=1}^{\infty} \binom{2n}{n} (H_{2n-1}- H_n)\frac{z^n}{n} = \log^2 \left(\frac{1+\sqrt{1-4z}}{2}\right) = \log^2 (C(z))}$$

Alternating version of Au-Yeung series: $\displaystyle \sum\limits_{n=1}^{\infty} (-1)^{n-1}\frac{H_n^2}{n^2}$

Problem: Closed form of $\displaystyle \sum\limits_{n=1}^{\infty} (-1)^{n-1}\frac{H_n^2}{n^2}$

We have seen how to deal with the Au-Yeung Series in three different ways in Closed form of Enrico Au-Yeung Series. However alternating versions (A) of Euler Sums often turn out to be trickier to deal with than the (H) counterpart.

Solution: We begin with the key identity:
                           
$$\begin{align}\sum\limits_{k=0}^{n} (-1)^k\binom{n}{k}\frac{H_{k+1}^{(3)}}{k+1} &= \frac{1}{n+1}\sum_{k=0}^{n} (-1)^k \binom{n+1}{k+1}\sum\limits_{j=0}^{k}\frac{1}{(j+1)^3}\tag{1}\\ &= \frac{1}{n+1}\sum_{j=0}^{n}\frac{1}{(j+1)^3} \sum\limits_{k=j}^{n}(-1)^k \binom{n+1}{k+1}\tag{2}\\ &= \frac{1}{n+1}\sum_{j=0}^{n}(-1)^{j} \binom{n}{j}\frac{1}{(j+1)^3}\tag{3} \\ &= \frac{1}{(n+1)^2}\sum_{j=0}^{n}(-1)^{j} \binom{n+1}{j+1}\frac{1}{(j+1)^2} \tag{4}\\ &= \frac{1}{(n+1)^2}\sum_{j=0}^{n}(-1)^{j} \frac{1}{(j+1)^2}\sum\limits_{k=j}^{n}\binom{k}{j}\tag{5}\\&= \frac{1}{(n+1)^2}\sum_{k=0}^{n} \sum_{j=0}^{k} (-1)^j\binom{k}{j}\frac{1}{(j+1)^2} \tag{6}\\&= \frac{1}{(n+1)^2}\sum_{k=0}^{n} \frac{1}{k+1}\sum_{j=0}^{k} (-1)^j\binom{k+1}{j+1}\frac{1}{j+1}\tag{7}\\&= \frac{1}{(n+1)^2}\sum_{k=0}^{n} \frac{1}{k+1}H_{k+1}\tag{8}\\ &= \frac{H_{n+1}^2+H_{n+1}^{(2)}}{2(n+1)^2}\tag{9}\end{align}$$

Justifications:
(1) Used $\displaystyle \binom{n+1}{k+1} = \frac{n+1}{k+1}\binom{n}{k}$ and applied change in order of summation.
(2) Used the identity: $\displaystyle \sum\limits_{k=j}^{n} (-1)^k\binom{n+1}{k+1} = \sum\limits_{k=j}^{n-1} (-1)^k\left(\binom{n}{k}+\binom{n}{k+1}\right) + (-1)^{n}\binom{n+1}{n+1} = (-1)^j\binom{n}{j}$ via telescoping summation.
(3) Used identity from (1).
(4) Used the identity: $\displaystyle \binom{n+1}{j+1} =  \sum\limits_{k=j}^{n} \binom{k}{j}$
(5) Interchanged order of summation.
(6) Reused identity from (1).
(7) Used the identity: $\displaystyle \sum\limits_{j=0}^{n} (-1)^j\binom{n+1}{j+1}\frac{1}{j+1} = H_{n+1}$.
(8) Used the identity: $\displaystyle \sum\limits_{k=1}^{n} \frac{H_k}{k} = \frac{H_n^2 + H_n^{(2)}}{2}$.

More generally, for a sequence of numbers $\{a_k\}_{k\ge 0}$, we denote $\displaystyle b_n = \sum\limits_{k=0}^{n} (-1)^k \binom{n}{k} a_k$ and the binomial inversion formula tell us that $\displaystyle a_n = \sum\limits_{k=0}^{n} (-1)^k \binom{n}{k} b_k$. Denoting $\displaystyle s_n = \sum\limits_{k=0}^{n} a_k$ we have,

\begin{align*}
\sum\limits_{k=0}^{n} (-1)^k \binom{n}{k} \frac{s_k}{k+1} &= \frac{1}{n+1}\sum\limits_{k=0}^{n} (-1)^k \binom{n+1}{k+1} \sum\limits_{j=0}^{k} a_j \\&= \frac{1}{n+1}\sum\limits_{j=0}^{n} a_j \sum\limits_{k=j}^{n} (-1)^k \binom{n+1}{k+1} \\&= \frac{1}{n+1}\sum\limits_{j=0}^{n} a_j \sum\limits_{k=j}^{n} (-1)^k \left[\binom{n}{k} + \binom{n}{k+1}\right] \\&= \frac{1}{n+1}\sum\limits_{j=0}^{n} (-1)^j \binom{n}{j} a_j = \frac{b_n}{n+1}
\end{align*}

Let us denote, $\displaystyle c_{n,r} = \sum\limits_{j_1 = 1}^{n}\sum\limits_{j_2 = 1}^{j_1} \cdots \sum\limits_{j_r = 1}^{j_{r-1}} \frac{1}{j_1 \cdots j_r}$ then we have, $\displaystyle c_{n,r} = \sum\limits_{j=1}^{n} \frac{c_{j,r-1}}{j} = c_{n-1,r} + \frac{c_{n,r-1}}{n}$ and consequently,

\begin{align*}
\sum\limits_{k=1}^{n} (-1)^{k-1} \binom{n}{k} c_{k,r} &= \sum\limits_{k=1}^{n} (-1)^{k-1} \left[\binom{n-1}{k-1} + \binom{n-1}{k} \right] c_{k,r} \\&= \sum\limits_{k=1}^{n} (-1)^{k-1} \binom{n-1}{k-1}  \left[c_{k-1,r} + \frac{c_{k,r-1}}{k}\right] + \sum\limits_{k=1}^{n-1} (-1)^{k-1} \binom{n-1}{k}  c_{k,r} \\&= \sum\limits_{k=1}^{n} (-1)^{k-1} \binom{n-1}{k-1}  c_{k-1,r} + \sum\limits_{k=1}^{n} (-1)^{k-1}\binom{n-1}{k-1}\frac{c_{k,r-1}}{k} + \sum\limits_{k=1}^{n-1} (-1)^{k-1} \binom{n-1}{k}  c_{k,r} \\&= \sum\limits_{k=1}^{n-1} (-1)^{k} \binom{n-1}{k}  c_{k,r} + \frac{1}{n}\sum\limits_{k=1}^{n} (-1)^{k-1} \binom{n}{k}c_{k,r-1} + \sum\limits_{k=1}^{n-1} (-1)^{k-1} \binom{n-1}{k}  c_{k,r} \\&= \frac{1}{n}\sum\limits_{k=1}^{n} (-1)^{k-1} \binom{n}{k}c_{k,r-1}
\end{align*}

Therefore, by $r$ times application we have, $\displaystyle \sum\limits_{k=1}^{n} (-1)^{k-1} \binom{n}{k} c_{k,r} = \frac{1}{n^r}$ and from binomial inversion result we have, $\displaystyle \sum\limits_{k=1}^{n} (-1)^{k-1} \binom{n}{k} \frac{1}{k^r} = c_{n,r}$.

Then, using $\displaystyle a_k = \frac{1}{(k+1)^r}$ and denoting by classical notation $\displaystyle s_k = H_{k+1}^{(r)}$ we have,

\begin{align*}
\sum\limits_{k=0}^{n} (-1)^k \binom{n}{k} \frac{H_{k+1}^{(r)}}{k+1} &= \frac{1}{n+1}\sum\limits_{j=0}^{n} (-1)^j \binom{n}{j} \frac{1}{(j+1)^r} \\&= \frac{1}{(n+1)^2}\sum\limits_{j=1}^{n+1} (-1)^{j-1} \binom{n+1}{j} \frac{1}{j^{r-1}} = \frac{c_{n+1,r-1}}{(n+1)^2}
\end{align*}

Our particular case being $r=3$, we have, $$\sum\limits_{k=0}^{n} (-1)^k \binom{n}{k} \frac{H_{k+1}^{(3)}}{k+1} = \frac{c_{n+1,2}}{(n+1)^2} = \frac{H_{n+1}^2 + H_{n+1}^{(2)}}{2(n+1)^2}$$

The Binomial Inversion formula tells us:

$$\sum\limits_{k=0}^{n} (-1)^k\binom{n}{k}\frac{H_{k+1}^2+H_{k+1}^{(2)}}{2(k+1)^2} = \frac{H_{n+1}^{(3)}}{n+1}$$

Now at this point the Euler accelaration formula(or the Euler Series Transformation) tells us:

$$\sum\limits_{n=0}^{\infty} (-1)^{n}\frac{H_{n+1}^2+H_{n+1}^{(2)}}{2(n+1)^2} = \sum\limits_{n=0}^{\infty} \frac{H_{n+1}^{(3)}}{2^{n+1}(n+1)}$$

That is after reindexing the summation: $$\sum\limits_{n=1}^{\infty} (-1)^{n-1}\frac{H_n^2}{n^2} + \sum\limits_{n=1}^{\infty} (-1)^{n-1}\frac{H_n^{(2)}}{n^2} = 2\sum\limits_{n=1}^{\infty}\frac{H_n^{(3)}}{2^nn}$$

The series in the RHS can be evaluated by observing that $\displaystyle \sum_{n=1}^{\infty}H_n^{(3)}x^n = \frac{\operatorname{Li}_3(x)}{1-x}$

Hence, $$\displaystyle \sum_{n=1}^{\infty}\frac{H_n^{(3)}}{n}x^{n} = \operatorname{Li}_4(x) -\frac{1}{2}\operatorname{Li}_2^2(x) - \log (1-x) \operatorname{Li}_3(x)$$

That is at $x = \dfrac{1}{2}$ we have: $$\displaystyle \sum\limits_{n=1}^{\infty} (-1)^{n-1}\frac{H_n^2}{n^2} + \sum\limits_{n=1}^{\infty} (-1)^{n-1}\frac{H_n^{(2)}}{n^2} =  2\sum\limits_{n=1}^{\infty} \frac{H_n^{(3)}}{2^{n}n} = 2\log 2 \operatorname{Li}_3\left(\frac{1}{2}\right) - \operatorname{Li}_2^2\left(\frac{1}{2}\right) + 2\operatorname{Li}_4\left(\frac{1}{2}\right)$$

Now it remains to deal with the alternating part: $\displaystyle \sum\limits_{n=1}^{\infty} (-1)^{n-1}\frac{H_n^{(2)}}{n^2}$ which was derived by R. Sitaramachandrarao in this paper.

$$\displaystyle  \sum\limits_{n=1}^{\infty} (-1)^{n-1}\frac{H_n^{(2)}}{n^2} = -\frac{17}{480}\pi^4 +4 \text{Li}_4 \left(\frac{1}{2} \right)+\frac{7}{2}\log(2) \zeta(3)-\frac{\pi^2 \log^2(2)}{6}+\frac{\log^4(2)}{6}$$

An alternative approach of calculating the same is to observe that:

$$\begin{align} \sum_{n=1}^\infty(-1)^{n-1}\frac{H_n^{(2)}}{n^2} &=\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^4}+\sum_{n=1}^{\infty}(-1)^{n-1}\frac{H_{n-1}^{(2)}}{n^2}\\ &=\frac{7}{8}\zeta(4)+\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^2}\sum_{k=1}^{n-1}\frac1{k^2}\\ &=\frac{7}{8}\zeta(4)+\sum_{k=1}^{\infty}\sum_{n=k+1}^{\infty}\frac{(-1)^{n-1}}{n^2k^2}\\ &=\frac{7}{8}\zeta(4)+\sum_{k=1}^{\infty}\sum_{n=1}^{\infty}\frac{(-1)^{k+n-1}}{(k+n)^2k^2}\\ &=\frac{7}{8}\zeta(4)+\sum_{k=1}^{\infty}\sum_{n=1}^{\infty}(-1)^{k+n-1}\left(\frac1{k^2n(n+k)}-\frac1{kn(k+n)^2}\right)\\ &=\frac{7}{8}\zeta(4)-\sum_{k=1}^{\infty}\sum_{n=1}^{\infty}\frac{(-1)^{k+n}}{k^2n(n+k)}+\sum_{k=1}^{\infty}\sum_{n=1}^{\infty}\frac{(-1)^{k+n}}{kn(k+n)^2}\\ &=\frac{7}{8}\zeta(4)-\frac{1}{2}\sum_{k=1}^{\infty}\sum_{n=1}^{\infty}(-1)^{k+n}\left(\frac{1}{kn^2(n+k)}+\frac{1}{k^2n(n+k)}\right)+\sum_{k=1}^{\infty}\sum_{n=1}^{\infty}\frac{(-1)^{k+n}}{kn(k+n)^2}\\&= \frac{7}{8}\zeta(4)-\frac{1}{2}\sum_{k=1}^{\infty}\sum_{n=1}^{\infty}\frac{(-1)^{k+n}}{n^2k^2}+\sum_{k=1}^{\infty}\sum_{n=1}^{\infty}\frac{(-1)^{k+n}}{kn(k+n)^2}\\&=\frac{7}{8}\zeta(4)-\frac{1}{8}\zeta^2(2)+\sum_{k=1}^{\infty}\sum_{n=1}^{\infty}\frac{(-1)^{k+n}}{kn(k+n)^2} \\ &= \frac{21}{8}\zeta(4)-\frac{1}{8}\zeta^2(2)+2\sum_{k=1}^{\infty}\frac{(-1)^kH_k}{k^3}\end{align}$$

The later alternating series evaluates to: $$\sum_{k = 1}^{\infty} \frac{(-1)^k}{k^3}H_k = -\frac{11\pi^4}{360}+\frac{\ln^42-\pi^2\ln^22}{12}+2\mathrm{Li}_4\left(\frac12\right)+\frac{7\ln 2}{4}\zeta(3)$$

which can be computed by expressing it as a logarithmic integrals. User Chris'ssis had shown me in the past an interesting way of dealing with the integrals.

Update: We have the more general relation, $$\displaystyle \sum\limits_{n=1}^{\infty} \dfrac{H_n}{n^3}x^n = -\frac{1}{2}\sum\limits_{n=1}^{\infty}\frac{1}{n^2}\sum\limits_{k=1}^{n}\frac{(1-x)^k}{k^2} -\frac{1}{2} \zeta(2)\operatorname{Li}_2(x)+\frac{7}{8}\zeta(4)-\frac{1}{4}\operatorname{Li}_2^2(1-x)+\frac{1}{4}\zeta^2(2)+\operatorname{Li}_4(x)\\+\frac{1}{4}\log^2 x\log^2(1-x)+\frac{1}{2}\log x\log (1-x)\operatorname{Li}_2(1-x)+\zeta(3)\log x -\log x\operatorname{Li}_2(1-x)$$

where, we may write $\displaystyle \sum\limits_{n=1}^{\infty}\frac{1}{n^2}\sum\limits_{k=1}^{n}\frac{(1-x)^k}{k^2} = \zeta(2)\operatorname{Li}_2(1-x) + \operatorname{Li}_4(1-x) - \sum\limits_{n=1}^{\infty} \frac{H_n^{(2)}}{n^2}(1-x)^n$.

I am too lazy to add a derivation for this. It's quite long and tedious for a blog post.

For those who are interested in similar logarithmic integrals can refer to Lewin's book Polylogarithms and associated functions pg. 310 formulae no. (10) to (12)). I don't have an elementary approach to this alternating Euler Sum.

Evaluation of an Interesting limit: #(53) form Wolfram Catalans Constant

To Display the LaTeX follow this link

Problem: The limit due to Glaisher (1877) referred as (53) in wolfram page of Catalan's Contant(G):

$\displaystyle exp\left(-\frac{1}{2}+\frac{2G}{\pi}\right) = \lim\limits_{n \to \infty} \frac{1}{(4n+1)^{2n}}\prod\limits_{k=1}^{n}\frac{(4k-1)^{4k-1}}{(4k-3)^{4k-3}}$

Link: Math.Se

Solution: We start with the integral representation of Catalan's constant:

$\displaystyle G = \frac{1}{2}\int\limits_0^{\frac{\pi}{2}} \dfrac{x}{\sin x}\,dx = \frac{\pi^2}{2}\int\limits_{0}^{\frac{1}{2}}\frac{x}{\sin \pi x}\,dx$

and we have, from partial fraction expansion of $\displaystyle \frac{\pi}{\sin x}$ ,

$$\begin{align}\frac{\pi}{\sin \pi x} &= \frac{1}{x} - 2x\sum\limits_{n=1}^{\infty} \frac{(-1)^n}{n^2 - x^2}\\

&= \sum\limits_{n=0}^{\infty} \int_0^1 (-1)^nt^{n-x}\,dt + (-1)^n t^{n-1+x} \,dt\\

&= \int_0^1 \sum\limits_{n=0}^{\infty} (-1)^nt^{n-x} \,dt + \int_0^1 \sum\limits_{n=0}^{\infty}(-1)^nt^{n-1+x} \,dt\\

&= \int_0^1 \frac{1}{t^x(1+t)}\,dt + \int_0^1 \frac{1}{t^{1-x}(1+t)} \\

&=\int_0^1 \frac{1}{t^x(1+t)}\,dt +\int_1^{\infty} \frac{1}{t^x(1+t)}\,dt \\

&= \int_0^{\infty} \frac{1}{t^x(1+t)}\,dt\end{align}$$

Then, $\displaystyle G = \frac{\pi}{2}\int\limits_{0}^{\frac{1}{2}} \int_0^{\infty} \frac{x}{t^x(1+t)}\,dt \,dx = \frac{\pi}{2}\int_0^{\infty} \int\limits_{0}^{\frac{1}{2}} \frac{x}{t^x(1+t)}\,dx\,dt$

Now, $\displaystyle \int\limits_{0}^{\frac{1}{2}} \frac{x}{t^x(1+t)}\,dx = \frac{1}{t+1}\left(\frac{1}{\log^2 t} - \frac{1}{\sqrt{t}\log^ t} - \frac{1}{2\sqrt{t}\log t} \right)$

Therefore, $\displaystyle \frac{2G}{\pi} = I = \int_0^{\infty} \frac{1}{t+1}\left(\frac{1}{\log^2 t} - \frac{1}{\sqrt{t}\log^ t} - \frac{1}{2\sqrt{t}\log t} \right)\,dt$

Making the substitiution, $t = e^{-w}$, the above integral becomes:

$$\begin{align} I &= \int_{-\infty}^{\infty} \frac{1}{w^2(1+e^{-w})}\left(e^{-w} - e^{-w/2} + \frac{w}{2}e^{-w/2}\right)\,dw\\

&=\int_{0}^{\infty} \frac{1}{w^2(1+e^{-w})}\left(e^{-w} - e^{-w/2} + \frac{w}{2}e^{-w/2}\right)\,dw+\int_{0}^{\infty} \frac{1}{w^2(1+e^{-w})}\left(1 - e^{-w/2} - \frac{w}{2}e^{-w/2}\right)\,dw\\

&=\int_{0}^{\infty} \frac{1}{w^2(1+e^{-w})}\left(1+ e^{-w} - 2e^{-w/2}\right)\,dw\end{align}$$

Introducing regulating factor $s$, we write:

$$I(s) = \int_0^{\infty} \frac{w^{s-2}}{1+e^{-w}}\left( 1+e^{-w} -2e^{-w/2} \right)\,dw$$

Therefore, $$\begin{align} I(s) &= \sum\limits_{n=0}^{\infty} (-1)^n \int_{0}^{\infty}\left( 1+e^{-w} -2e^{-w/2} \right)w^{s-2}e^{-nw}\,dw\\

&=\sum\limits_{n=0}^{\infty} (-1)^n\left(\frac{\Gamma(s-1)}{n^{s-1}} + \frac{\Gamma(s-1)}{(n+1)^{s-1}} -\frac{2^s\Gamma(s-1)}{(2n+1)^{s-1}}\right)\end{align}$$

Using the fact that, $\displaystyle \lim\limits_{s \to 0} s\Gamma(s-1) = -1$, we have,

$$I = \lim\limits_{s \to 0} \sum\limits_{n=0}^{\infty} \frac{(-1)^{n+1}}{s}\left(\frac{1}{n^{s-1}} + \frac{1}{(n+1)^{s-1}} - \frac{2^s}{(2n+1)^{s-1}}\right)$$

Since, $$\begin{align} \lim_{s \to 0} \frac{\frac{1}{n^{s-1}} + \frac{1}{(n+1)^{s-1}} - \frac{2^s}{(2n+1)^{s-1}}}{s} &= n\left(2\log(n+1/2)-\log(n+1)-\log(n)\right) \\&+ \log(n+1/2)-\log(n+1)\end{align}$$

We have, $$\displaystyle \begin{align} I &= \sum_{n=0}^{\infty} (-1)^{n+1} n\left(2\log(n+1/2)-\log(n+1)-\log(n) \right) \\&+ \sum_{n=0}^{\infty} (-1)^{n+1} \left(\log(n+1/2)-\log(n+1)\right)\\

&= \frac{1}{2} + \lim\limits_{n \to \infty} \log \frac{1}{(4n+1)^{2n}} \prod_{k=0}^{n-1}\frac{(4k+3)^{4k+3}}{(4k+1)^{4k+1}}\end{align}$$

Thus, $$\displaystyle \exp{\frac{2G}{\pi} - \frac{1}{2}} = \lim\limits_{n \to \infty} \frac{1}{(4n+1)^{2n}} \prod\limits_{k=0}^{n-1}\frac{(4k+3)^{4k+3}}{(4k+1)^{4k+1}}$$