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I need to prove that the limit in this bountied question $$\lim_{n\to \infty} \frac{n\ 4^{2n}}{e^{2n}}\{d_{2n}b_n\}\leq \frac{3}{4}$$ where $d_{2n}=\text{LCM}(1,2,...,2n)$ and $\{x\}$ is the fractional part of $x$. It is known by the above link that $$b_n:= -\sum_{j=0}^n\binom nj^2(2(H_{n-j}-H_j)\ln((j+n)!)$$

I would really appreciate an answer which I can accept. Thank you!

Max
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  • What motivates this problem? Could you give us some context? – Gary Aug 03 '23 at 11:38
  • @Gary Thanks for your comment. I was just thinking about this problem. Nothing particular other than the link given in the question. Mathematica gives that after some $N$ we have the $$\frac{n\ 4^{2n}}{e^{2n}} \left{e^{2n}\binom{2n}{n}\left(\log \left( {\frac{{3n}}{2}} \right) +\mathcal{O}!\left( {\frac{1}{n}} \right)\right)\right}\to 0 $$ – Max Aug 03 '23 at 11:42
  • Shouldn’t $a_n$ contain $\binom{2n}n\gamma$ or a $\sum\limits_{j=0}^n \binom n j^2 H_{j+n}$? – Тyma Gaidash Aug 03 '23 at 12:06
  • @TymaGaidash Please see the question linked in this post. $a_n$ is the sequence as defined in the linked post and $$b_n=-a_n= -\sum_{j=0}^n\binom nj^2(2(H_{n-j}-H_j)\ln((j+n)!) $$ and by the link in this post Gary proved that as $n\to \infty$ $$b_n=-a_n=\binom{2n}{n}\left(\log \left( {\frac{{3n}}{2}} \right) +\mathcal{O}!\left( {\frac{1}{n}} \right)\right) $$ – Max Aug 03 '23 at 12:19
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    Using the $O(1/n)$ formulation won't be productive, because it's being multiplied by $e^{2n}$ (and then some) which completely scrambles the values of the fractional part. – Greg Martin Aug 03 '23 at 17:10
  • @GregMartin Then can we just use the asymptotic with the $O(1/n)$ part? Can the another representation of $b_n$ be used? I request you to write as an answer. – Max Aug 03 '23 at 17:33

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