# Dorin Marghidanu's Believe It Or Not

### Solution 1

It's well known that,

$\displaystyle\lim_{n\to\infty}E_{n}=\left(1+\frac{1}{n}\right)^{n}=e.$

If $G_n=H_n-\ln n,\,$ another Euler's sequence then $\displaystyle\lim_{n\to\infty}G_n=\gamma.\,$ Now, employing the Stolz-Cesaro Lemma,

\displaystyle\begin{align} \lim_{n\to\infty}\left(E_{n+1}^{H_{n+1}}-E_n^{H_n}\right) &= \lim_{n\to\infty}\frac{E_{n+1}^{H_{n+1}}-E_n^{H_n}}{(n+1)-n}=\lim_{n\to\infty}\frac{E_n^{H_n}}{n}\\ &=\lim_{n\to\infty}\frac{E_n^{G_n+\ln n}}{n}=\lim_{n\to\infty}E_n^{G_n}\frac{E_n^{\ln n}}{n}\\ &=\lim_{n\to\infty}E_n^{G_n}\lim_{n\to\infty}\frac{E_n^{\ln n}}{n}=e^{\gamma}\cdot 1\\ &=e^{\gamma}. \end{align}

### Solution 2

We have that

$\displaystyle \lim_{n\rightarrow +\infty}E_{n+1}^{H_{n+1}}-E_{n}^{H_{n}}=\lim_{n\rightarrow +\infty }\frac{ E_{n+1}^{H_{n+1}}-E_{n}^{H_{n}}}{n+1-n}$

so, by the Stolz-Cesàro Lemma,

$\displaystyle \lim_{n\rightarrow +\infty}E_{n+1}^{H_{n+1}}-E_{n}^{H_{n}}=\lim_{n\rightarrow +\infty }\frac{E_{n}^{H_{n}}}{n}.$

Now,

\begin{align} \ln (\frac{E_{n}^{H_{n}}}{n}) &=H_{n}\ln E_{n}-\ln n \\ &=H_{n}n\ln (1+\frac{1}{n})-\ln n \\ &=(\ln n+\gamma +O(\frac{1}{n}))n(\frac{1}{n}+O(\frac{1}{n^{2}}))-\ln n \\ &=(\ln n+\gamma +O(\frac{1}{n}))(1+O(\frac{1}{n}))-\ln n \\ &=\gamma +O(\frac{1}{n}) \end{align}

so

$\displaystyle\lim_{n\rightarrow +\infty }\ln (\frac{E_{n}^{H_{n}}}{n})=\gamma.$

and we have done.

### Acknowledgment

Dorin Marghidanu has kindly posted that beautiful problem at the CutTheKnotMath facebook page. Solution 1 is by Dorin Marghidanu (and, independently, Diego Alvariz); Solution 2 by Giulio Francot.