# How to calculate $\lim_{n\to\infty}(1+1/n^2)(1+2/n^2)\cdots(1+n/n^2)$?

How can we compute the following limit:

$$\lim_{n\to\infty}(1+1/n^2)(1+2/n^2)\cdots(1+n/n^2)$$

Mathematica gives the answer $\sqrt{e}$. However, I do not know to do it.

• is it $1+\left(\frac 1n\right)^2$ instead of $1+\frac 1{n^2}?$ – lab bhattacharjee May 12 '13 at 6:35
• @labbhattacharjee What is the difference? – user17762 May 12 '13 at 6:35
• @user17762, I mean for each term is it $1+\left(\frac rn\right)^2$ instead of $1+\frac r{n^2}?$ – lab bhattacharjee May 12 '13 at 6:36
• @labbhattacharjee I think it is $1+\dfrac{r}{n^2}$, since it gives the limit as $\sqrt{e}$. Also, the product will diverge, it it were to be $1+\dfrac{r^2}{n^2}$. – user17762 May 12 '13 at 6:37

We will estimate $$f(n) = \sum_{k=1}^n \log \left(1 + \dfrac{k}{n^2}\right)$$ Recall that $\log(1+x) = x + \mathcal{O}(x^2)$. Hence, we get that $$f(n) = \sum_{k=1}^n \left(\dfrac{k}{n^2} + \mathcal{O}\left(\dfrac{k^2}{n^4}\right)\right) = \dfrac{n+1}{2n} + \mathcal{O}(1/n) \tag{\star}$$ where we made use of the fact that $\displaystyle \sum_{k=1}^n k = \dfrac{n(n+1)}2$ and $\displaystyle \sum_{k=1}^n k^2 = \dfrac{n(n+1)(2n+1)}6$.

Now letting $n \to \infty$ in $(\star)$, we get that $$\lim_{n \to \infty} f(n) = \dfrac12$$ The product you are interested in is $e^{f(n)}$ and since $e^x$ is continuous, we have $$\lim_{n \to \infty} e^{f(n)} = e^{\lim_{n \to \infty} f(n)} = e^{1/2} = \sqrt{e}$$


• Nice approach +1. my previous deleted comment was due to the fact that sometimes the overline used in rising factorial is not visible on screen. – Paramanand Singh Apr 8 '17 at 5:21
• @ParamanandSingh Thanks. Maybe, you have some problem with your video. There's a $\texttt{\overline}$ there. I just increased it. – Felix Marin Apr 8 '17 at 5:23
• @ParamanandSingh I just checked it. Sometimes it's not visible after we reboot the page. If the problem persists I'll change the notation. I like this notation because it's very intuitive. It 'shows' that it's increasing. – Felix Marin Apr 8 '17 at 5:27
• Thanks for the extra effort. It is now visible. the earlier version was visible only when I used zoom feature of mathjax. Surprisingly the mobile browser on android is better and it is able to display your earlier version too. Problem is probably my low resolution laptop. – Paramanand Singh Apr 8 '17 at 5:27
• I don't get the first equality.. what does $\overline n$ mean? – Ant Dec 27 '17 at 16:43

For $1\le k\le n$, we have

\begin{align} \log\left(1+{k\over n^2}\right)-{k\over n(n+1)} &=\int_1^{1+k/n^2}{dt\over t}-{k\over n(n+1)}\\ &=\int_0^{k/n^2}\left({1\over1+t}-{n\over n+1}\right)dt\\ &={1\over n+1}\int_0^{k/n^2}\left(1-tn\over1+t \right)dt\\ &\le{1\over n+1}\int_0^{k/n^2}dt\\ &\le{k\over n^2(n+1)}\\ &\le{1\over n(n+1)} \end{align}

Note, the integral in the third line is clearly non-negative, which means the initial expression is also non-negative. Therefore

$$0\le\log\left(1+{k\over n^2}\right)-{k\over n(n+1)}\le{1\over n(n+1)}$$

Consequently, since $1+2+\cdots+n={n(n+1)\over2}$, we have

$$0\le\sum_{k=1}^n\log\left(1+{k\over n^2}\right)-{1\over 2}=\sum_{k=1}^n\left(\log\left(1+{k\over n^2}\right)-{k\over n(n+1)}\right)\le{1\over n+1}$$

so by the Squeeze Theorem

$$\lim_{n\to\infty}\sum_{k=1}^n\log\left(1+{k\over n^2}\right)={1\over2}$$

and thus

$$\lim_{n\to\infty}(1+1/n^2)(1+2/n^2)\cdots(1+n/n^2)=e^{1/2}=\sqrt e$$

Using

$$\lim_{n\to\infty}\left(1+\frac an\right)^n=e^a$$

We have

\begin{align} L&=\lim_{n\to\infty} \left[\left((1+\frac{1}{n^2})(1+\frac{2}{n^2})(1+\frac{3}{n^2})\cdots(1+\frac {n} {n^2})\right)^{n^2}\right]^{\frac1{n^2}}\\ &=\lim_{n\to\infty}e^{\frac{1+\ldots+n}{n^2}}=\lim_{n\to\infty}e^\frac{n(n+1)}{2n^2}=\sqrt e \end{align}

• This is wrong since the fact that, for every fixed $k$, $$\lim_{n\to\infty}x_{k,n}=x_k$$ does not imply in general that $$\lim_{n\to\infty}\left(\prod_{k=1}^nx_{k,n}\right)^{n^2}=\lim_{n\to\infty}\left(\prod_{k=1}^nx_k\right)^{n^2}$$ – Did Dec 31 '17 at 16:23