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Some times ago, In a mathematical problem book I sow that this formula. I don't no whether it is true or not. But now I'm try to prove it. I have no idea how to begin it. Any hint or reference would be appreciate. Thank you.

$$\lim_{n \to \infty} \prod_{k=1}^n \left (1+\frac {kx}{n^2} \right) =e^{x/2}$$

StubbornAtom
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Bumblebee
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2 Answers2

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Notice that $x-x^2/2\leq \ln(1+x) \leq x$ for $x\geq 0$ then : $$\frac{xk }{n^2} \geq \ln\left(1+\frac{ xk}{n^2}\right) \geq \frac{x k}{n^2} -\frac{x^2 k^2}{n^4} $$ Sum form $k=1$ to $n$ : $$\frac{(n+1)x}{2n} \geq \sum_{k=1}^n \ln\left(1+\frac{ xk}{n^2}\right) \geq \frac{(n+1)x}{2n} - \frac{(n+1)(2n+1)x}{6n^3} $$ Then the limit of the middle sum is $x/2$ take the exponential to get that the requested limit is $e^{x/2}$.

Bumblebee
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Tulip
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Here I use the method suggested by Daniel Fischer in the comment section:
Take the logarithm of $$\prod _{k=1}^n \left(1+\dfrac{kx}{n^2} \right).$$ Then we have $$\ln\left(\prod_{k=1}^n\ \left(1+\dfrac{kx}{n^2 }\right) \right)=\sum_{k=1}^n\ln\left(1+\dfrac{kx}{n^2 }\right).$$ By the Taylor series expansion of logarithms, $$ \ln\left(1+\dfrac{kx}{n^2} \right)=\sum_{m=1}^\infty\dfrac{(-1)^{m-1}}{m}\left(\dfrac{ kx}{n^2 }\right)^m \\= \left(\dfrac{kx}{n^2 }\right)-\dfrac{1}{2} \left(\dfrac{kx}{n^2 }\right)^2+\dfrac{1}{3} \left(\dfrac{kx}{n^2 }\right)^3-\dfrac{1}{4} \left(\dfrac{kx}{n^2 }\right)^4+\dfrac{1}{5} \left(\dfrac{kx}{n^2 }\right)^5-\dfrac{1}{6} \left(\dfrac{kx}{n^2 }\right)^6+ …$$ Now $$\sum_{k=1}^n\ln\left(1+\dfrac{kx}{n^2} \right) =\dfrac{x}{n^2 }\left( \sum_{k=1}^nk\right)-\dfrac{x^2}{2n^4 }\left( \sum_{k=1}^nk^2\right)+\dfrac{x^3}{3n^6 } \left(\sum_{k=1}^nk^3 \right)-\dfrac{x^4}{4n^8 }\left(\sum_{k=1}^nk^4 \right)+ …$$ Note that $$\sum_{k=1}^nk^r =\dfrac{n^{r+1}}{r+1}+\dfrac{n^r}{2} + O(n^{r-1})$$ is a $(r+1)^{\text{th}}$ degree polynomial of $n.$
Therefore $$\lim_{n\to\infty}\left(\dfrac{1}{n^{2r}} \sum_{k=1}^nk^r\right)=\left\{ \begin{array}{rl} \frac{1}{2}:&r=1\\ 0:&r>1 \end{array} \right. $$ Hence we can obtain $$\lim_{n\to\infty}\ln\left(\prod_{k=1}^n\left(1+\dfrac{kx}{n^2 }\right) \right)=\lim_{n\to\infty}\sum_{k=1}^n\ln\left(1+\dfrac{kx}{n^2 }\right)=\dfrac{x}{2}-0+0-0+ ...$$ $$\ln \left(\lim_{n\to\infty}\prod_{k=1}^n\left(1+\dfrac{kx}{n^2 }\right)\right)=\dfrac{x}{2}$$ $$\lim_{n\to\infty}\prod_{k=1}^n\left(1+\dfrac{kx}{n^2 }\right)=e^{x/2}$$

Bumblebee
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