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$$ L = \lim_{n \to \infty}\prod_{i = 1}^{n}\left(\, 1 + {i \over n^{2}}\,\right) $$

Consider, $$ L_{1} = \prod_{i = 1}^{n}\lim_{n \to \infty}\left(\,1 + {i \over n^{2}}\,\right) $$ Q1. Is $L = L_{1}$ ?. Probably not but why can't I distribute the limit over product ?.

Q2. As I see it, each term in $L_{1}$ is individually $1$, therefore $L_{1} = 1\cdot 1\cdots \infty$ times $ = 1$. But if I write $L_{2} = \lim_{n \to \infty}\left(\, 1 + {1 \over n}\,\right)^{n}$ as $$ \lim_{n \to \infty}\left(\,1 + {1 \over n}\,\right)\cdot \lim_{n \to \infty}\left(\, 1 + {1 \over n}\,\right)\cdots \infty\ \mbox{times,} $$ I get $1$ again because each term goes to $1$. However, it is equal to $\,\mathrm{e}$. The mistake lies in the distribution of limit part. What is the mistake ?.

Please be elaborate because I can't get it ?.

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    $\begingroup$ Notice that $\text{L}=\sqrt{e}$ and $\text{L}_1=\prod_{i=1}^{n}1=1$. So $\text{L}\ne\text{L}_1$ $\endgroup$ Oct 29, 2016 at 17:20

2 Answers 2

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$$\prod_{i=1}^{n}\left(1+\frac{i}{n^2}\right)=\exp\sum_{i=1}^{n}\log\left(1+\frac{i}{n^2}\right) $$ and the function $f(x)=\log(1+x)$ is bounded between $x$ and $x-\frac{x^2}{2}$ on the interval $[0,1]$, hence $$\sum_{i=1}^{n}\log\left(1+\frac{i}{n^2}\right)=o(1)+\frac{1}{n^2}\sum_{i=1}^{n}i = \frac{1}{2}+o(1) $$ as $n\to +\infty$, and the wanted limit equals $\exp\left(\frac{1}{2}\right)=\color{red}{\large\sqrt{e}}$.

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In the formula $$L_1=\prod_{i=1}^{\style{color:#f00}{n}}\lim_{n\to \infty}(a+\frac i {n^2})$$ the $n$ marked with red does not denote anything, it is neither bound to an expression nor is it predefined, whereas in the definition of $L$ is is bounded by the $\lim_{n\to \infty}$. It is the same as if you would take $$\prod_{i=0}^1\prod_{j=1}^ij$$ and write $$\prod_{j=1}^{\style{color:red}{i}}\prod_{i=0}^1j$$ There is simply no rule that enables you to even state that they are equal, not alone proving that they are equal.

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