Calculate the limit $\lim_{n\rightarrow\infty}\left(\frac{\prod_1^n(1 +r/n^2)}{e^{1/2}}\right)^n$

Question

How to calculate the limit $$\lim_{n\rightarrow\infty}\left(\frac{\prod\limits_{r=1}^n\left(1 +\frac{r}{n^2}\right)}{e^{1/2}}\right)^n?$$ I tried using the sandwich theorem by taking logarithm of the limit and using the expansion

$$ x - (x^2)/2<\ln(1+x)<x$$ but I was getting different from answer from the two sides. The correct answer is $e^{1/3}$.

Answer 1

Using the fact that $\log(1+x)=x-\frac{x^2}{2}+\mathcal{O}(x^3)$ for small $x$, we find \begin{align*} \left( {\prod\limits_{r = 1}^n {\left( {1 + \frac{r}{{n^2 }}} \right)} } \right)^n & = \exp \left( {n\sum\limits_{r = 1}^n {\log \left( {1 + \frac{r}{{n^2 }}} \right)} } \right) \\ &= \exp \left( {n\left( {\sum\limits_{r = 1}^n {\left( {\frac{r}{{n^2 }} - \frac{1}{2}\frac{{r^2 }}{{n^4 }} + \mathcal{O}\!\left( {\frac{{r^3 }}{{n^6 }}} \right)} \right)} } \right)} \right) \\ & = \exp \left( {\frac{1}{n}\sum\limits_{r = 1}^n r - \frac{1}{{2n^3 }}\sum\limits_{r = 1}^n {r^2 } + \mathcal{O}\!\left( {\frac{1}{{n^5 }}} \right)\sum\limits_{r = 1}^n {r^3 } } \right) \\ &= \exp \left( {\frac{{n + 1}}{2} - \frac{{(n + 1)(2n + 1)}}{{12n^2 }} + \mathcal{O}\!\left( {\frac{1}{n}} \right)} \right) \\ & = \exp \left( {\frac{n}{2} + \frac{1}{3} + \mathcal{O}\!\left( {\frac{1}{n}} \right)} \right). \end{align*} The result now follows.

Answer 2

Hint: If we take

$\begin{align} &P(x) := \prod_{r=1}^n\left(x +\frac{r}{n^2}\right) \\[1mm] &= x^n + \frac{\sum_{r_1=1}^n r_1}{n^2}x^{n-1} + \frac{\sum_{r_1=1}^n\sum_{r_2=1}^n r_1r_2}{n^4}x^{n-2} + \dots + \frac{\sum_{r_1=1}^n\sum_{r_2=1}^n\dots\sum_{r_n=1}^n r_1r_2\cdot...\cdot r_n}{n^{2n}} \\[1mm] &= x^n + \frac{\sum_{r_1=1}^n r_1}{n^2}x^{n-1} + \frac{\left(\sum_{r_1=1}^n r_1\right)^2}{n^4}x^{n-2} + \dots + \frac{\left(\sum_{r_1=1}^n r_1\right)^n}{n^{2n}} \\[1mm] &= x^{n} + \frac{S}{n^2}x^{n-1} + \frac{S^2}{n^4}x^{n-2} + \dots + \frac{S^n}{n^{2n}} \tag{$\star$} \end{align}$

where $$S = \sum_{r_1=1}^n r_1 = \frac{n(n+1)}{2}$$

Then the numerator of the ratio of the limit $$\lim_{n\rightarrow\infty}\left(\frac{\prod_{r=1}^n(1 +r/n^2)}{e^{1/2}}\right)^n$$

will be

$$\begin{align} P(1) &= 1 + \frac{S}{n^2} + \frac{S^2}{n^4} + \dots + \frac{S^n}{n^{2n}} \\[1mm] &= \frac{1 - \frac{S^{n+1}}{n^{2(n+1)}}}{1 - \frac{S}{n^2}} \\[1mm] &= \frac{n\left(-n^{-n-1}\left(n+1\right)^{n+1}+2^{n+1}\right)}{2^n\left(n-1\right)}\\[1mm] &= \frac{n}{n-1}\frac{\left(2^{n+1} - \left(1+\frac{1}{n}\right)^{n+1}\right)}{2^n} \\[1mm] &\sim 2 - \frac{1}{2^n}\left(1+\frac{1}{n}\right)^{n}\end{align}$$

Answer 3

You can squeeze as follows:

As you already realized you may use the Taylor expansion: $$\ln (1+x) = x-\frac 1{2(1+\xi_x)^2}x^2 \text{ for some } \xi_x \text{ with } 0 < \xi_x <x <1$$

Setting

$$s_n := n\sum_{r=1}^n\ln\left(1+\frac r{n^2} \right)- \frac n2$$

you get the estimates (using on the RHS that $\frac r{n^2} \leq \color{blue}{\frac 1n} $)

$$n\sum_{r=1}^n\left(\frac r{n^2}-\frac 1{2}\frac{r^2}{n^4}\right)-\frac n2\leq s_n \leq n\sum_{r=1}^n\left(\frac r{n^2}-\frac 1{2(1+\color{blue}{\frac 1n})^2}\frac{r^2}{n^4}\right)-\frac n2$$ Now some rearranging and using $\sum_{r=1}^n r^2 = \frac{n(n+1)(2n+1)}{6}$ gives

$$\frac 12 - \frac{(n+1)(2n+1)}{12n^2}\leq s_n \leq \frac 12 - \frac{(n+1)(2n+1)}{12n^2\left(1+\frac 1n\right)^2}$$

Taking the limits gives the desired result $\frac 12 - \frac 16 = \frac 13$.