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So while trying to solve a limit problem I ran into this expression $$ \lim_{n\to \infty} [(n+1) e^{{1 \over n+1} (\sum_{r=1}^{n+1}\ln({r\over n+1}))} - (n) e^{{1 \over n} (\sum_{r=1}^{n}\ln({r\over n}))} ].$$

What I want to know is that if this is reimann summable or solvable in any way?

It would help if you could justify your answer rigorously.

Paramanand Singh
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Hrishabh Nayal
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2 Answers2

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Let's observe that $$f(n) \to\exp\left(\int_{0}^{1}\log x\, dx\right) =\frac{1}{e}$$ and expression under limit in question can be written as $$f(n+1)+n\{f(n+1)-f(n)\}$$ whose limit is same as that of $$\frac{1}{e}+nf(n)\frac{x_n-1}{\log x_n} \cdot\log x_n\tag{1}$$ where $$x_n=\frac{f(n+1)}{f(n)}\to 1$$ Thus the desired limit crucially depends on the limit of $$n\{\log f(n+1)-\log f(n)\} \tag{2}$$ and one can prove that the above tends to $0$ so that the desired limit is $1/e$.

We can write $(2)$ as $$\left(1-\frac{1}{n+1}\right)\sum_{k=1}^{n+1}\log\frac{k}{n+1}-\sum_{k=1}^{n}\log\frac{k}{n}$$ or $$\log\frac{(n+1)!}{(n+1)^{n+1}}-\log\frac{n!}{n^n}-\frac{1}{n+1}\sum_{k=1}^{n+1}\log\frac{k}{n+1}$$ which equals $$n\log\frac{n}{n+1}-\log f(n+1)$$ The above clearly tends to $-1-(-1)=0$ and our job is done.

On the other hand if we look closely at your $f(n) $ we see that $$nf(n) =\sqrt[n] {n!} $$ and thus your question is essentially the same as this one. Why do ask the same question twice in two seemingly different forms? At the same time I don't think my answer is fundamentally different from the answer given there by user medicu.

Paramanand Singh
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  • I asked the question there but I could not get answers that would either build up on my approach or would prove it wrong. So I asked a different question on my approach. Is this against site rules? – Hrishabh Nayal Mar 31 '20 at 02:46
  • @HrishabhNayal: no, it's not against the rules, but I don't see anything extra gained here. Also I am not annoyed, just wanted to know the reason you asked. You should put a link of the older question in your current question. – Paramanand Singh Mar 31 '20 at 02:55
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Considering $$a_p=p\, e^{{1 \over p} \sum_{r=1}^{p}\log\left({r\over p}\right)}$$ we have, using Pochhammer symbols $$\sum_{r=1}^{p}\log\left({r\over p}\right)=\log \left(\left(\frac{1}{p}\right)^p (1)_p\right)$$ which makes $$a_p=\left((1)_p\right){}^{\frac{1}{p}}\implies \log(a_p)={\frac{1}{p}}\log\left((1)_p\right)$$ Using Stirling approximations for large $p$ $$\log\left((1)_p\right)=p (\log (p)-1)+\frac{1}{2} \log (2 \pi p)+\frac{1}{12 p}+O\left(\frac{1}{p^3}\right)$$ $$\log(a_p)=\log (p)-1+\frac{\log (2 \pi p)}{2 p}+\frac{1}{12 p^2}+O\left(\frac{1}{p^4}\right)$$ $$a_p=e^{\log(a_p)}=\frac{p}{e}+\frac{\log (2 \pi p)}{2 e}+\frac{3 \log ^2(2 \pi p)+2}{24 e p}+O\left(\frac{1}{p^2}\right)$$ Apply it twice and continue with Taylor expansions $$a_{n+1}-a_n=\frac{1}{e}+\frac{1}{2 e n}+O\left(\frac{1}{n^2}\right)$$

You could do the same with factorials since $$\sum_{r=1}^{p}\log\left({r\over p}\right)=\sum_{r=1}^{p}\log(r)-\sum_{r=1}^{p}\log(p)=\log(p!)-p\log(p)$$

Claude Leibovici
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