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I'm sure you guys can do this in many ways, using integrals, Taylor series, but I need a
high school way, like the use of a simple squeeze theorem. Can we get such a way?

$$\lim_{n\to\infty}\left(1+\log\left(1+\frac{1}{n^2}\right)\right)\left(1+\log\left(1+\frac{2}{n^2}\right)\right)\cdots\left(1+\log\left(1+\frac{n}{n^2}\right)\right)$$

EDIT: The proof I need I plan to present to some kids, so everything should be clear for them.

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Approach 1

We can use the inequality $$ x-\frac12x^2\le\log(1+x)\le x\tag{1} $$ for $x\ge0$, to good result.

First, note that $$ \begin{align} \color{#0000FF}{\sum_{k=1}^n\frac{k}{n^2}} &=\frac{n^2+n}{2n^2}\\ &=\frac12+\frac1{2n}\tag{2} \end{align} $$ Next, since $\frac{k}{n^2}\le\frac1n$, we get $$ \begin{align} \color{#0000FF}{\sum_{k=1}^n\frac{k}{n^2}}-\frac1{2n} &=\sum_{k=1}^n\left[\frac{k}{n^2}-\frac1{2n^2}\right]\\ &\le\color{#00A000}{\sum_{k=1}^n\log\left(1+\frac{k}{n^2}\right)}\\ &\le\color{#0000FF}{\sum_{k=1}^n\frac{k}{n^2}}\tag{3} \end{align} $$ Finally, each of the terms $\log\left(1+\frac{k}{n^2}\right)$ is less that $\frac1n$, so we get $$ \begin{align} \color{#00A000}{\sum_{k=1}^n\log\left(1+\frac{k}{n^2}\right)}-\frac1{2n} &=\sum_{k=1}^n\left[\log\left(1+\frac{k}{n^2}\right)-\frac1{2n^2}\right]\\ &\le\color{#C00000}{\sum_{k=1}^n\log\left(1+\log\left(1+\frac{k}{n^2}\right)\right)}\\ &\le\color{#00A000}{\sum_{k=1}^n\log\left(1+\frac{k}{n^2}\right)}\tag{4} \end{align} $$ Putting together $(2)$, $(3)$, and $(4)$ implies $$ \frac12-\frac1{2n}\le\color{#C00000}{\sum_{k=1}^n\log\left(1+\log\left(1+\frac{k}{n^2}\right)\right)}\le\frac12+\frac1{2n}\tag{5} $$ An application of the Squeeze Theorem yields $$ \lim_{n\to\infty}\sum_{k=1}^n\log\left(1+\log\left(1+\frac{k}{n^2}\right)\right)=\frac12\tag{6} $$ and therefore, $$ \lim_{n\to\infty}\prod_{k=1}^n\left(1+\log\left(1+\frac{k}{n^2}\right)\right)=e^{1/2}\tag{7} $$


Approach 2

Here we only assume $\log(1+x)\lt x$ for all $x\gt-1$. Then, because $1-\frac{x}{1+x}=\frac1{1+x}$, $$ \frac{x}{1+x}\le-\log\left(1-\frac{x}{1+x}\right)=\log(1+x)\le x\tag{8} $$ Then, for $0\le x\le\frac1n$, $(8)$ implies $$ \frac{n}{n+1}x\le\frac{x}{1+x}\le\log(1+x)\le x\tag{9} $$ Applying $(9)$ twice shows that for $0\le x\le\frac1n$, $$ \left(\frac{n}{n+1}\right)^2x\le\log(1+\log(1+x))\le x\tag{10} $$ Summing $(10)$, we get $$ \left(\frac{n}{n+1}\right)^2\sum_{k=1}^n\frac{k}{n^2}\le\sum_{k=1}^n\log\left(1+\log\left(1+\frac{k}{n^2}\right)\right)\le\sum_{k=1}^n\frac{k}{n^2}\tag{11} $$ Applying $(2)$ gives $$ \frac12\frac{n}{n+1}\le\sum_{k=1}^n\log\left(1+\log\left(1+\frac{k}{n^2}\right)\right)\le\frac12\frac{n+1}{n}\tag{12} $$ An application of the Squeeze Theorem yields $$ \lim_{n\to\infty}\sum_{k=1}^n\log\left(1+\log\left(1+\frac{k}{n^2}\right)\right)=\frac12\tag{13} $$ and therefore, $$ \lim_{n\to\infty}\prod_{k=1}^n\left(1+\log\left(1+\frac{k}{n^2}\right)\right)=e^{1/2}\tag{14} $$

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    $\begingroup$ AWESOME (+1) $\endgroup$ – user 1591719 Nov 9 '14 at 18:49
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I think use this inequality can solve it:

use this follow well know inequality: $$\dfrac{x}{x+1}\le ln{(1+x)}\le x,x>0$$ so $$\dfrac{x}{x+1}\le \ln{(1+\ln{(1+x)})}\le x$$ $$\dfrac{i}{n^2+n}\le \dfrac{i}{n^2+i}\le\ln{(1+\ln{(1+\dfrac{i}{n^2})})}\le\dfrac{i}{n^2}$$ so $$\sqrt{e}=e^{\displaystyle\lim_{n\to\infty}\sum_{i=1}^{n}\dfrac{i}{n^2+n}}\le I=e^{\displaystyle\lim_{n\to\infty}\sum_{i=1}^{n}\ln{(1+\ln{(1+\dfrac{i}{n})})}}<e^{\displaystyle\lim_{n\to\infty}\sum_{i=1}^{n}\dfrac{i}{n^2}}=e^{1/2}$$ this limits is $\sqrt{e}$

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Lemma: For two triangular array of real-numbers $\{s_{m,n}\}$ and $\{r_{m,n}\}$, where, $n \in \mathbb{N}$ and $m = 1,2,3,\cdots,n$, ($r_{m,n} \neq 0$) if, $\dfrac{s_{m,n}}{r_{m,n}} \to 1$, an $n \to \infty$ uniformly with respect to $m$, that is for an $\epsilon >0$, $\exists\,N_0$, s.t., $\left|\dfrac{s_{m,n}}{r_{m,n}} - 1\right| < \epsilon$, for $\forall n \ge N_0$ and $m = 1,2,3,\cdots,n$, then $$\displaystyle \lim\limits_{n \to \infty} \sum\limits_{m=1}^{n} r_{m,n} = \lim\limits_{n \to \infty} \sum\limits_{m=1}^{n} s_{m,n} \textrm{ (if exists) }$$

Proof: Write $\displaystyle \dfrac{s_{m,n}}{r_{m,n}} = 1+\epsilon_{m,n}$. Then $\epsilon_{m,n} \to 1$, an $n \to \infty$ uniformly with respect to $m$.

Thus, $\displaystyle \sum\limits_{m=1}^{n} s_{m,n} = \sum\limits_{m=1}^{n} r_{m,n} + \sum\limits_{m=1}^{n} \epsilon_{m,n}r_{m,n}$

Since, $\displaystyle \lim\limits_{n \to \infty} \sum\limits_{m=1}^{n} r_{m,n}$ exists, choose $M>0$ such that $\displaystyle \left|\sum\limits_{m=1}^{n} r_{m,n}\right| < M$, $\forall n \in \mathbb{N}$

Moreover, $|\epsilon_{m,n}| < \frac{\epsilon}{M}$, for $m = 1,2,\cdots,n$, for sufficiently large $n$. Hence, $\displaystyle \left|\sum\limits_{m=1}^{n} \epsilon_{m,n}r_{m,n}\right| \le \epsilon$, for sufficiently large $n$.

Thus, $\displaystyle \lim\limits_{n \to \infty} \sum\limits_{m=1}^{n} \epsilon_{m,n}r_{m,n} = 0$, proving our claim.

Now, for any positive sequence $\{x_n\}$ which converges to $0$, we have $\dfrac{\log(1+x_n)}{x_n} \to 1$,

applying lemma twice we have $\displaystyle \lim\limits_{n\to \infty} \sum\limits_{m=1}^{n} \log\left(1+\log\left(1+\frac{m}{n^2}\right)\right) = \lim\limits_{n\to \infty} \sum\limits_{m=1}^{n} \log\left(1+\frac{m}{n^2}\right) = \lim\limits_{n\to \infty} \sum\limits_{m=1}^{n} \frac{m}{n^2} = \frac{1}{2}$

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