# Interesting Identity: $\lim_{m\to\infty}\lim_{n\to\infty}\prod_{k=1}^{\infty}\prod_{j=2}^{m}(1+\phi_{n})^{nk^{-j}\Phi_{j}}=e^{\gamma}$

Define the following identities: $$\phi_{n}:=\frac{4}{\pi}\int_{0}^{\infty}\frac{\coth(nx^{-1})-xn^{-1}}{n(1+x^{2})^{2}}\;dx \qquad \text{and}\qquad \Phi_{n}:=\frac{\cos(n\pi)}{n}$$ $$\forall n\in\mathbb{N}$$. I want to show that : $$\lim_{m\to\infty}\lim_{n\to\infty}\prod_{k=1}^{\infty}\prod_{j=2}^{m}(1+\phi_{n})^{ nk^{-j}\Phi_{j}}=e^{\gamma}$$ Where $$\gamma$$ is the Euler - Mascheroni constant.

What I did is that I used the linearity of integrals for $$\phi_{n}$$ $$\frac{\pi}{4}\phi_{n}=\int_{0}^{\infty}\frac{\coth\left(\frac{n}{x}\right)}{n(1+x^{2})^{2}}\;dx-\int_{0}^{\infty}\frac{x}{n^{2}(1+x^{2})^{2}}\;dx$$ The first one I define it as $$I_{1}$$ and the second one I define it as $$I_{2}$$. Afterwards I defined: $$\lambda_{n}(x):=\frac{\coth\left(\frac{n}{x}\right)}{n(1+x^{2})^{2}}\leq\left|\frac{1}{n(1+x^{2})^{2}}\right|\leq\left|\frac{1}{2n(1+x^{2})}\right|=:V_{n}(x)$$ so that: $$\lim_{n\to\infty}\int_{0}^{\infty}V_{n}(x)\;dx=\lim_{n\to\infty}\frac{\pi}{4n}=0$$ Applying the Lebesgue Dominated Convergence Theorem I get: $$\lim_{n\to\infty}\int_{0}^{\infty}\lambda_{n}(x)\;dx=0 \tag 1$$ Now note that: $$I_{2}=\int_{0}^{\infty}\frac{x}{n^{2}(1+x^{2})}\;dx=\frac{1}{2n^{2}}\quad\implies\quad\lim_{n\to\infty}I_{2}=0 \tag 2$$ $$(1)$$ and $$(2)$$ verify that $$\lim_{n\to\infty}(1+\phi_{n})^{n}\leq\lim_{n\to\infty}(1+I_{1}-\displaystyle I_{2})^{n}=\lim_{n\to\infty}\left(1+\frac{1}{n}-\frac{2}{n\pi^{2}}\right)^{n}=e$$

Now all I have to evaluate $$\displaystyle\exp\left(\lim_{m\to\infty}\sum_{k=1}^{\infty}\sum_{j=2}^{\infty}\frac{\cos(j\pi)}{k^{j}j}\right)$$ but this seems a bit confusing and misleading when I try it which is what I am stuck on.

• Your direction is right; you made some mistakes when estimating $\phi_n$ (namely: $\coth(\ldots)\color{red}{\geqslant}1$; an inequality is insufficient to compute the needed exact value of $\lim\limits_{n\to\infty}n\phi_n$, not just an estimate). Commented Jan 27, 2021 at 9:13
• You are correct I apologize. Attention: I updated the proof because I did not define $V_{n}(x)$ properly. Commented Jan 27, 2021 at 23:08

The expression under the limit(s) is equal to $${a_n}^{b_m}$$, where $$a_n=(1+\phi_n)^n,\qquad b_m=\sum_{k=1}^\infty\sum_{j=2}^m\frac{(-1)^j}{jk^j}.$$
Let $$f(t)=\coth(1/t)-t$$, then $$n\phi_n=(4/\pi)\int_0^\infty(1+x^2)^{-2}f(x/n)\,dx$$. Observe that $$f$$ is decreasing (to check, take the derivative and use $$\sinh z>z$$ for $$z>0$$). Also, $$f(0)=1$$ if understood as the limit.
Thus, $$n\mapsto f(x/n)$$ is increasing for each fixed $$x$$, so DCT (or even MCT) is applicable, and $$\lim\limits_{n\to\infty}n\phi_n=(4/\pi)\int_0^\infty(1+x^2)^{-2}f(0)\,dx=1$$, hence $$\lim\limits_{n\to\infty}a_n=e$$ as you know.
As for $$b_m$$, the corresponding $$\sum_{k=\color{red}{2}}^\infty\sum_{j=2}^{\color{red}{\infty}}$$ is absolutely convergent, so that we're free to interchange the summations, hence to take $$m\to\infty$$ directly under the outer sum: $$\lim_{m\to\infty}b_m=\sum_{k=1}^\infty\sum_{j=2}^\infty\frac{(-1/k)^j}{j}=\sum_{k=1}^\infty\left[\frac1k-\log\left(1+\frac1k\right)\right]\\=\lim_{n\to\infty}\sum_{k=1}^n[\ldots]=\lim_{n\to\infty}\left[\left(\sum_{k=1}^n\frac1k\right)-\log(n+1)\right].$$ As $$\lim\limits_{n\to\infty}\big(\log(n+1)-\log n\big)=0$$, we then have $$\lim\limits_{m\to\infty}b_m=\gamma$$ "almost by definition".
Finally, the given limit is $$e^\gamma$$ because $$(a,b)\mapsto a^b$$ is continuous at $$(e,\gamma)$$.