13
$\begingroup$

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.

$\endgroup$
2
  • 1
    $\begingroup$ 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). $\endgroup$
    – metamorphy
    Commented Jan 27, 2021 at 9:13
  • $\begingroup$ You are correct I apologize. Attention: I updated the proof because I did not define $V_{n}(x)$ properly. $\endgroup$ Commented Jan 27, 2021 at 23:08

1 Answer 1

4
$\begingroup$

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 $(e,\gamma)$.

$\endgroup$
1
  • $\begingroup$ Thank you very much for completing the proof! $\endgroup$ Commented Jan 27, 2021 at 23:09

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .