# Thread: Proving equivalency (limits)

1. ## Proving equivalency (limits)

There are given $\displaystyle x>0$ real number and $\displaystyle 0<x_{n,1}, x_{n,2}, ...,x_{n,n}$ real numbers and $\displaystyle x_{n,1}+ x_{n,2}+ ...+ x_{n,n}=x$ for all $\displaystyle n$. Prove that:
$\displaystyle \displaystyle\lim_{n \to \infty}\prod_{i=1}^{n}(1+{x_{n,i}})=e^x \Longleftrightarrow \lim_{n \to \infty}\displaystyle \max_{1\leq i\leq n}x_{n,i}=0$.

Hint: Consider $\displaystyle \log(1+y)-y$ function near $\displaystyle y=0$.

Thank you for your help in advance.
Ps: Happy $\displaystyle \pi$ Day!

2. Originally Posted by zadir
There are given $\displaystyle x>0$ real number and $\displaystyle 0<x_{n,1}, x_{n,2}, ...,x_{n,n}$ real numbers and $\displaystyle x_{n,1}+ x_{n,2}+ ...+ x_{n,n}=x$ for all $\displaystyle n$. Prove that:
$\displaystyle \displaystyle\lim_{n \to \infty}\prod_{i=1}^{n}(1+{x_{n,i}})=e^x \Longleftrightarrow \lim_{n \to \infty}\displaystyle \max_{1\leq i\leq n}x_{n,i}=0$.

Hint: Consider $\displaystyle \log(1+y)-y$ function near $\displaystyle y=0$.
The terms of the Taylor series of $\displaystyle \log(1+y)$ (for y>0) alternate in sign. It follows that $\displaystyle y-\frac12y^2<\log(1+y)<y$. Therefore

$\displaystyle \displaystyle\sum_{i=1}^{n}(x_{n,i} - \tfrac12x_{n,i}^2) < \sum_{i=1}^{n} \log(1+x_{n,i}) < \sum_{i=1}^{n}x_{n,i} = x.$

Take exponentials to see that $\displaystyle \displaystyle e^xe^{-\frac12\sum_i x_{n,i}^2} < \prod_{i=1}^{n} (1+x_{n,i}) < e^x.$

So to prove the result, it suffices to show that $\displaystyle \displaystyle \lim_{n\to\infty}\sum_{i=1}^n x_{n,i}^2 = 0\; \Longleftrightarrow\; \lim_{n \to \infty} \max_{1\leqslant i\leqslant n}\{x_{n,i}\}=0.$

In one direction, this is fairly obvious: if the limit on the right is not zero then the greatest term in the sum on the left will not tend to 0, hence neither will the sum. For the other direction, notice that

$\displaystyle \displaystyle \sum_{i=1}^n x_{n,i}^2 \leqslant \max_{1\leqslant i\leqslant n}\{x_{n,i}\}\sum_{i=1}^n x_{n,i}.$

Thus if $\displaystyle \max_i x_{n,i}\leqslant\varepsilon$, it follows that $\displaystyle \sum_i x_{n,i}^2 \leqslant \varepsilon x.$