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lim $\frac{1}{n}[(m+1)(m+2)...(m+n)]^\frac{1}{n}$ as n tends to infinity

I took y = lim $[(m+1)(m+2)...(m+n)]^\frac{1}{n}$

Taking log both sides

log y = lim $\frac{1}{n}*log[(m+1)(m+2)...(m+n)]$

which gives lim $\frac{1}{n}\sum_{1}^{n} log(m+k)$ (n tends to infinity)

This could be written in integral form as $\int_{0}^{1}log(m+x) dx$

I will use Ilate rule after this step, however, I am not sure if I substituted x correctly above. In other similar questions, substituting k/n with x helped solving them. Please help me proceed.

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    It cannot work in this way, right from the beginning ; you have as well to take the log of $1/n$ ... – Jean Marie Jun 23 '24 at 09:16
  • @JeanMarie I will substitute the value and divide by n as pointed out by you (maybe LHopital rule will work). But first I want to integrate the expression shown above. –  Jun 23 '24 at 09:24
  • "lim $\frac{1}{n}\sum_{1}^{n} log(m+k)$ (n tends to infinity) could be written in integral form as $\int_{0}^{1}log(m+x) dx$": ??? – Anne Bauval Jun 23 '24 at 10:27

1 Answers1

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I think you forgot $\log(1/n)$.

$$\begin{split} \log y & = \lim_{n\to\infty}\frac{1}{n}\log\left((m+1)\dots(m+n)/n^n\right) \\ & = \lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^n\log\left(\frac{m}{n}+\frac{k}{n}\right) \end{split}$$ and then proceed with the integration.

L0wc3ll
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