Limits of a product

Question

$N$ is a positive integer. I want to calculate this limit, but i couldn't get anywhere when i tried.

$$P[n]=\left(1+\frac {1}{n^2}\right)\left(1+\frac {2}{n^2}\right)\cdots\left(1+\frac {n-1}{n^2}\right)$$ as $n\to \infty$

I tried to apply $\ln()$ at both sides to transform into a sum etc.. tried to use functions to limit the superior and inferior intervals of the function like this: $$\exp\left(\left(n-1\right)\ln\left(1+\frac1{n^2}\right)\right)<P[n]<\exp\left((n-1)\ln\left(1+\frac{n-1}{n^2}\right)\right)$$ The only thing I get is that: $1<\lim(P[n])<e$

Can anyone help me to solve this?

Answer 1

$$\left(1+\frac {j}{n^2}\right)\left(1+\frac {n-j}{n^2}\right)=1+\frac{1}{n}+\frac{j(n-j)}{n^4}$$

By AM-GM we have $$\sqrt{j(n-j)}\leq \frac{n}{2} $$ and hence $$1+\frac{1}{n} \leq \left(1+\frac {j}{n^2}\right)\left(1+\frac {n-j}{n^2}\right) \leq 1+\frac{1}{n}+\frac{1}{4n^2}$$

Therefore $$\left(1+\frac{1}{n}\right)^{\frac{n-1}{2}} \leq P[n] \leq \left( 1+\frac{1}{n}+\frac{1}{4n^2} \right)^\frac{n}{2}$$

Now $$\lim_n \left(1+\frac{1}{n}\right)^{\frac{n-1}{2}}=\left( \left(1+\frac{1}{n}\right)^{n} \right)^{\frac{n-1}{2n}}=\sqrt{e}$$ And $$\lim_n \left( 1+\frac{1}{n}+\frac{1}{4n^2} \right)^\frac{n}{2} =\lim_n \left(\left( 1+\frac{4n+1}{4n^2} \right)^\frac{4n^2}{4n+1}\right)^\frac{4n+1}{8n} =\sqrt{e}$$

Therefore $$\lim_n P[n]=\sqrt{e}$$

Answer 2

Let $x$ be a real number in the interval $[0,1/2)$. Then we have the following known inequality: $$ 1+x \le \exp x\ .$$ (Which holds for any real $x$.) This can be used to give an upper bound for the product, while also giving the chance to convert in the formula for the bound to a sum, which can be easily computed. We want a similar argument for a lower bound. For this, note that $$ \frac 1{1+x}\le 1-x+x^2\le \exp(-x+x^2) = \frac 1{\exp(x-x^2)}\ . $$ (We have applied the same inequality above for an other argument.) From the double inequality: $$ \exp(x-x^2)\le 1+x\le \exp(x) $$ we get \begin{align} \exp\sum_{0<k<n}\left(\frac k{n^2}-\frac {k^2}{n^4}\right) &= \prod_{0<k<n}\exp\left(\frac k{n^2}-\frac {k^2}{n^4}\right) \\\\ &\le \prod_{0<k<n}\left( 1+\frac k{n^2}\right) \\\\ &\le \prod_{0<k<n}\exp\frac k{n^2} \\\\ &= \exp\sum_{0<k<n}\frac k{n^2} \ . \end{align} The expressions at the beginning and at the end enclose the given sequence and converge each to $\exp\frac 12$, which is thus also the limit of the given sequence.

Edit: Computer check using sage:

sage: R = RealField( 400 )
sage: N = 1000
sage: print "P[%s] ~ %.11f" % ( N, prod( [1+k/N^2 for k in [1..N-1] ] ) ) 
P[1000] ~ 1.64762303820
sage: print "exp(1/2) ~ %.11f" % R(exp(1/2))
exp(1/2) ~ 1.64872127070

Answer 3

First realize that, $$\prod_{i=1}^{n} 1+ \frac i{n^2} = \prod_{i=1}^n \frac {n^2+i}{n^2} = \prod_{i=n^2+1}^{n^2+n} \frac {i}{n^2} $$

Then,

$$\prod_{i=n^2+1}^{n^2+n} \frac {i}{n^2} = \exp \left ( \sum_{i=n^2+1}^{n^2+n} \log(i) - 2n\log(n) \right )$$

Then we use a Riemann inequality since $\log$ is an increasing function,

$$\int_{n^2}^{n^2+n} \log(x)dx \leq \sum_{i=n^2+1}^{n^2+n} \log(i) \leq \int_{n^2+1}^{n^2+n+1} \log(x)dx \mathrm{\ \ (Riemann \ inequality)} $$ $$\begin{align} \int_{n^2}^{n^2+n} \log(x)dx -2n\log(n) & = [xlog(x)-x]_{n^2}^{n^2+n} -n\log(n^2) \\ & = (n^2+n) \log(n^2+n) -n^2 - n - n^2\log(n^2) +n^2 -n \log(n^2)\\ & = (n^2 +n) \log(1+\frac 1 n) - n \\ & = (n^2 +n)(\frac 1 n - \frac 1 {2n^2} + O(n^{-3})) - n \underset{n\to \infty}\to \frac 12 \end{align} $$

Likewise

$$\begin{align} \int_{n^2+1}^{n^2+n+1} \log(x)dx -2n\log(n) & = [x\log(x)-x]_{n^2+1}^{n^2+n+1} -n\log(n^2) \\ & = (n^2+1)\log(1+\frac{n}{n^2+1})+n\log(1+\frac{n+1}{n^2})-n \underset{n\to \infty}\to \frac 12 \end{align} $$

Therefore $\log(P_n) \underset{n\to \infty}\to \frac 12$ and $P_n \underset{n\to \infty}\to \sqrt e$

The only result I used is $x\mathcal \in V(0), \ \log(1+x)=x-\frac 12 x^2+O(x^3)$ to take the limits. Thanks to @achille hui for the correction.

Answer 4

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \prod_{k = 1}^{n - 1}\pars{1 + {k \over n^{2}}} & = {1 \over n^{2n - 2}}\prod_{k = 1}^{n - 1}\pars{k + n^{2}} = {1 \over n^{2n - 2}}\pars{1 + n^{2}}^{\overline{n - 1}} = {1 \over n^{2n - 2}} \,{\Gamma\pars{1 + n^{2} + n - 1} \over \Gamma\pars{1 + n^{2}}} \\[5mm] & \stackrel{\mrm{as}\ n\ \to\ \infty}{\sim}\,\,\, {1 \over n^{2n - 2}} \,{\root{2\pi}\pars{n + n^{2}}^{1/2 + n + n^{2}}\expo{-n - n^{2}} \over \root{2\pi}\pars{1 + n^{2}}^{3/2 + n^{2}}\expo{-1 - n^{2}}} \\[5mm] & = {1 \over n^{2n - 2}} \,{n^{1 + 2n + 2n^{2}}\pars{1 + 1/n}^{1/2 + n + n^{2}} \over n^{3 + 2n^{2}}\pars{1 + 1/n^{2}}^{3/2+ n^{2}}}\,\expo{-n + 1} \\[5mm] & \stackrel{\mrm{as}\ n\ \to\ \infty}{\sim}\,\,\, \,{\pars{1 + 1/n}^{n} \over \pars{1 + 1/n^{2}}^{n^{2}}}\,\bracks{\pars{1 + {1 \over n}}^{n^{2}}\expo{-n + 1}} \\[5mm] & \stackrel{\mrm{as}\ n\ \to\ \infty}{\Large \to}\,\,\, \lim_{n \to \infty}\exp\pars{n^{2}\ln\pars{1 + {1 \over n}} - n + 1} = \expo{\color{red}{1/2}} = \bbx{\root{\expo{}}} \approx 1.6487 \end{align}

Note that $\ds{n^{2}\ln\pars{1 + {1 \over n}} - n + 1 \,\,\,\stackrel{\mrm{as}\ n\ \to\ \infty}{=}\,\,\, \color{red}{1 \over 2} + {1 \over 3n} + \mrm{O}\pars{1 \over n^{2}}}$.

Answer 5

Considering $$P_n=\prod_{i=1}^{n-1} \left(1+\frac i {n^2} \right)\implies \log(P_n)=\sum_{i=1}^{n-1} \log\left(1+\frac i {n^2} \right)$$ Since $n$ is large, by Taylor (or even using simple equivalents) $$\log\left(1+\frac i {n^2} \right)=\frac{i}{n^2}-\frac{i^2}{2 n^4}+O\left(\frac{1}{n^6}\right)$$ making $$\log(P_n)=\frac{(n-1) \left(6 n^2-2 n+1\right)}{12 n^3}+\cdots=\frac{1}{2}-\frac{2}{3 n}+O\left(\frac{1}{n^2}\right)$$ showing a limit of $\frac 12$. Then $P_n \to \sqrt e$.

Probably too early, but added for your curiosity, using Pochhammer symbols $$P_n=\frac{\left(n^2+1\right){}_{n-1} } {n^{2n-2}}$$ Using asymptotics $$\log(P_n)=\frac{1}{2}-\frac{2}{3 n}+\frac{1}{3 n^2}+O\left(\frac{1}{n^3}\right)$$ Continuing with Taylor $$P_n=e^{\log(P_n)}=\sqrt e\left(1-\frac{2}{3 n}+\frac{5}{9 n^2}\right)+O\left(\frac{1}{n^3}\right)$$ which shows the limit and how it is approached.

It also allows a shortcut evaluation of the partial products even for small values of $n$ as shown below $$\left( \begin{array}{cccc} n & P_n & P_n \approx & \text{approximation} \\ 3 & \frac{110}{81} & 1.35802 & 1.38411 \\ 4 & \frac{2907}{2048} & 1.41943 & 1.43118 \\ 5 & \frac{570024}{390625} & 1.45926 & 1.46553 \\ 6 & \frac{1873495}{1259712} & 1.48724 & 1.49097 \\ 7 & \frac{20872566000}{13841287201} & 1.50799 & 1.51039 \\ 8 & \frac{418915572075}{274877906944} & 1.52401 & 1.52564 \\ 9 & \frac{316401132826240}{205891132094649} & 1.53674 & 1.53790 \end{array} \right)$$