Following is the well know result that connects Harmonic number with binomial coefficients, $$\color{red}{H_n}=\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k}{n\choose k}$$
This motivates me and I came to find the following identity for Euler-Mascheroni constant, $\gamma$.
$$\lim_{j\to \infty}\sum_{m=1}^{j}\left(\sum_{n=1}^{m}\sum_{k=1}^{n}(-1)^{k+1} \color{red}{H_k}{n\choose k}-\ln m\right)\frac{1}{j}=\gamma$$
My solution Due to Euler we have that $$H_k=\int_0^1\frac{1-x^k}{1-x}dx$$ and using the result we observe that $$\sum_{k=1}^{n}(-1)^{k+1} H_k{n\choose k}=\int_0^1\sum_{k=1}^n\left(\frac{1-x^k}{1-x}\right)\frac{(-1)^{k+1}}{1}{n\choose k}dx=\int_0^1\frac{1+(1-x)^k-1}{1-x}dx=\int_0^1 (1-x)^{n-1}dx =\frac{1}{n}$$ Here we make the use of result $$\small{ \sum_{\color{green}{k=0}}^{n}(-1)^k{n\choose k} =0,\;\;\sum_{\color{green}{k=0}}^n(-1)^{k+1}{n\choose k}x^{k}=(1-x)^n}$$ and thus we have $$\lim_{j\to\infty}\sum_{m=1}^{j}\left(\sum_{n=1}^m\frac{1}{n}-\ln m\right)\frac{1}{j}=\lim_{j\to\infty}\sum_{m=1}^{j}\left(H_m-\ln m\right)\frac{1}{j}\cdots(1)$$ By Stolz- Cesaro theorem result $(1)$ it directly follows $$\displaystyle \lim_{j\to \infty} (H_{j+1}-\ln(j+1))=\gamma$$ Result can be also be proved in the following manner. $$\lim_{j\to\infty}\sum_{m=1}^{j}\left(H_m-\ln m\right) =\lim_{j\to\infty}\sum_{m=1}^j\left(\sum_{i=1}^m\frac{1}{i}-\ln m \right)=\lim_{j\to\infty}\left(\sum_{m=1}^j\frac{j-m+1}{m}-\ln\ j!\right)\frac{1}{j}$$ Using the Stirling approximation of $\ln j!= j\ln j- j+O(\ln j)$ we have then $$\lim_{j\to \infty}\left(H_j-\ln j\right)=\gamma$$ As the claimed to found to be true. So, I'm curious to know if we can prove the result in different way/s than presentated here.How can we prove the result in others ways?
Thank you