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I am having trouble with the limit; $$\lim_{m\to\infty}\left(\sqrt[m]{m+1}\cdot\sqrt[m^2+m]{\binom{m}{1}\binom{m}{2}\cdots\binom{m}{m}}\right) .$$

where $$ \binom{m}{j}=\frac{m!}{(m-j)!j!}$$ I tried to use the Stirling formula but still I could not clarify the steps.

By the way Mathematica says the answer should be $\sqrt{e}$.

Any nice trick for this ?

Guy Fsone
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  • Possible duplicate of https://math.stackexchange.com/questions/2023222/limit-involving-binomial-coefficients-lim-n-to-infty-left-binomn0?rq=1 – Sangchul Lee Jan 15 '18 at 11:20

2 Answers2

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Note that

$$\sqrt[m]{m+1}=e^{\frac{\log(m+1)}{m}}\to e^0=1$$

and

$$\sqrt[m^2+m]{\binom{m}{1}\binom{m}{2}\cdots\binom{m}{m}}=e^{\frac{\log\left(\sum_{k=1}^m\binom{m}{k}\right)}{m^2+m}}\to\sqrt e$$

indeed by Stolz-Cesaro:

$$\lim_{m\to\infty}\frac{\sum_{k=1}^m\log\binom{m}{k}}{m^2+m}=\lim_{m\to\infty}\frac{\sum_{k=1}^{m+1}\log\binom{m+1}{k}-\sum_{k=1}^m\log\binom{m}{k}}{(m+1)^2+(m+1)-(m^2+m)}=\lim_{m\to\infty}\frac{\sum_{k=1}^m\log \frac{m+1}{m+1-k}}{2m+2}=\lim_{m\to\infty}\frac{\log \frac{(m+1)^m}{m!}}{2m+2}=\lim_{m\to\infty}\frac12\cdot \frac{m}{m+1}\cdot \log \frac{(m+1)}{\sqrt[m]{m!}}=\frac12\cdot 1 \cdot \log e=\frac12$$

user
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  • Very nice use of Cesaro-Stolz. The last limit (which evaluates to $e$) is also handled by Cesaro-Stolz. +1 – Paramanand Singh Jan 16 '18 at 15:13
  • @ParamanandSingh Thanks! for the last limit I had in mind the ratio-root criteria which is a particular case of SC as you have already pointed out earlier. – user Jan 16 '18 at 15:19
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Using the technique of double summation,

$$ \sum_{k=1}^{n} \log (k!) = \sum_{k=1}^{n} \sum_{j=1}^{k} \log j = \sum_{j=1}^{n} (n+1-j) \log j. $$

So it follows that

\begin{align*} \sum_{k=1}^{m} \log \binom{m}{k} &= (m+1) \log (m!) - 2 \sum_{k=1}^{m} \log (k!) \\ &= \sum_{j=1}^{m} (2j-m-1) \log j \\ &= \sum_{j=1}^{m} (2j-m-1) \log \left(\frac{j}{m}\right) \end{align*}

The last line follows from the fact that $\sum_{j=1}^{m} (2j-m-1) \log m = 0$. So we have

\begin{align*} \frac{1}{m(m+1)}\sum_{k=1}^{m} \log \binom{m}{k} &= \frac{m}{m+1} \sum_{j=1}^{m} \left(\frac{2j-1}{m}-1\right) \log \left(\frac{j}{m}\right) \, \frac{1}{m} \\ &\xrightarrow[m\to\infty]{} \int_{0}^{1} (2x-1)\log x \, dx = \frac{1}{2}. \end{align*}

(Although justifying this convergence requires more than the usual Riemann sum argument, it is still manageable.)

Together with $(m+1)^{1/m} \to 1$, we conclude that the limit is $e^{1/2} = \sqrt{e}$.

Sangchul Lee
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