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$$\lim_{n \to \infty} \frac{1}{n}\left((m+1)(m+2) \cdots (m+n)\right)^{\frac{1}{n}}$$

My try:$$A=\lim_{n \to \infty} \frac{1}{n}\left((m+1)(m+2) \cdots (m+n)\right)^{\frac{1}{n}}$$

Taking Natural log we get

$$\ln(A)=\lim_{n \to \infty}\frac{1}{n}\sum_{k=1}^n\ln\left(\frac{m+k}{n}\right)$$

But i cant use Riemann Sum , Any idea?

Umesh shankar
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2 Answers2

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$$A=\frac{1}{n}\big((m+1)(m+2) \cdots (m+n)\big)^{\frac{1}{n}}=\frac{1}{n}\left(\frac{\Gamma (m+n+1)}{\Gamma (m+1)}\right)^{\frac{1}{n}}$$ Take logarithms $$\log(nA)=\frac{1}{n}\big(\log (\Gamma (m+n+1))- \log (\Gamma (m+1)) \big)$$ Now, using the very first term of Stirling approximation for large $n$ you should get $$\log(nA)=-1+\log(n)+O\left(\frac{1}{n}\right)$$ Continuing with Taylor $$nA=e^{\log(nA)}=\frac n e +\cdots$$

Edit

If you want to the know the impact of $m$ on the result, yo need to add the next term in Stirling approximation and, using the same process, you should get $$nA=\frac n e +\frac 1 e \left(\left(m+\frac{1}{2}\right) \log (n)-\log (\Gamma (m+1))+\frac{1}{2} \log (2 \pi ) \right)+O\left(\frac{1}{n^2}\right) $$

Claude Leibovici
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You are almost there $$ \lim_{n\to\infty}\frac1n\sum_{k=1}^n\log\left(\frac{m+k}{n}\right)=\int_0^1\log(x)\,\mathrm{d}x $$ The limits are $\int_{m/n}^{1+m/n}$ as $n\to\infty$.

robjohn
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