Seeking the result: $\lim_{j,M \to \infty}\frac{1}{M}\prod_{k=1}^{M}\left[\prod_{n=j}^{2j}\left(1+\frac{1}{kn}\right)\right]^{\frac{1}{ln 2}}$

Question

I am seeking the result of this limit $$\lim_{j,M \to \infty}\frac{1}{M}\prod_{k=1}^{M}\left[\prod_{n=j}^{2j}\left(1+\frac{1}{kn}\right)\right]^{\frac{1}{ln 2}}=X$$

$X=1.78107...$ I have tried takes the log but it takes me no where.

How to determine the result of this limit?

Answer 1

\begin{align} \log\left(\frac{1}{M}\prod_{k=1}^{M}\left[\prod_{n=j}^{2j}\left(1+\frac{1}{kn}\right)\right]^{\frac{1}{\ln 2}}\right) &=-\log(M)+\frac{1}{\log 2}\sum_{k=1}^{M}\sum_{n=j}^{2j}\log\left(1+\frac{1}{kn}\right)\\ &=-\log(M)+\frac{1}{\log 2}\sum_{n=j}^{2j}\sum_{k=1}^{M}\left(\frac{1}{kn}+O\Bigl(\frac{1}{k^2n^2}\Bigr)\right)\\ &=-\log(M)+\frac{1}{\log 2}\sum_{n=j}^{2j}\frac 1n\sum_{k=1}^{M}\frac 1k+\frac{1}{\ln 2}\sum_{n=j}^{2j}\frac 1{n^2}\sum_{k=1}^{M}\frac 1{k^2}\\ &=-\log(M)+\frac{1}{\log 2}\left(\log(2)+O\Bigl(\frac 1j\Bigr)\right)\left(\log(M)+\gamma+O\Bigl(\frac 1M\Bigr)\right)+o(1)\\ &=-\log(M)+\left(1+O\Bigl(\frac 1j\Bigr)\right)\left(\log(M)+\gamma+O\Bigl(\frac 1M\Bigr)\right)+o(1)\\ &=\gamma+O\left(\frac{\log(M)}j\right)\\ &\to\gamma \end{align} hence your limit converges to $e^\gamma$, provided that $\log(M)/j\to 0$.