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Let $x, y \in \mathbb{R}$. Prove that $\lim_{n\rightarrow \infty} n(1-\sqrt{(1-\frac{x}{n})(1-\frac{y}{n})} )=\frac{x+y}{2}$

I tried some stuff but somehow I can not manage to find the right direction to prove that. Further it also required to state for which n, the elements of the series are defined, but isn't that dependent on the sign of x and y?

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    $\begingroup$ Try multiplying the expression by $1=\frac{1+\sqrt{(1-\frac{x}{n})(1-\frac{y}{n})}}{1+\sqrt{(1-\frac{x}{n})(1-\frac{y}{n})}}$ $\endgroup$ – MATHBOI Apr 19 at 15:06
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\begin{align}\lim_{n\to\infty}n\left(1-\sqrt{\left(1-\frac{x}{n}\right)\left(1-\frac{y}{n}\right)}\:\right)\:\ &=\lim_{n\to\infty}n\left(1-\sqrt{\left(1-\frac{x}{n}\right)\left(1-\frac{y}{n}\right)}\:\right)\frac{1+\sqrt{\left(1-\frac{x}{n}\right)\left(1-\frac{y}{n}\right)}}{1+\sqrt{\left(1-\frac{x}{n}\right)\left(1-\frac{y}{n}\right)}} \\ &=\lim_{n\to\infty}\frac{x+y-\frac{xy}{n}}{1+\sqrt{\left(1-\frac{x}{n}\right)\left(1-\frac{y}{n}\right)}}\\ &=\frac{x+y+0}{1+\sqrt{(1+0)(1+0)}}\\ &=\frac{x+y}{2} \end{align} Sorry for being lazy in the comments. I hope this helps haha!

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  • $\begingroup$ Perfect, thank you. could you also maybe help me with the second question, for which n the elements of the series are defined? $\endgroup$ – Parinn Apr 19 at 15:58
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    $\begingroup$ I believe it would be all $n\neq 0$ to avoid division by zero. Also, you would require $n\geq x$ to avoid imaginary components. Also, $(1+x/n)(1+y/n)>0$. You can solve this inequality for $n$. I believe that's all the restrictions on $n$ $\endgroup$ – MATHBOI Apr 19 at 16:03
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Hint:

For large enough $n:$

$(1-x/n)(1-y/n)=$

$1-(x+y)/n +O(1/n^2);$

$(1-(x+y)/n+O(1/n^2))^{1/2}= $

$1-(1/2)(x+y)/n +O(1/n^2);$

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You can use $\text{HM}\le\text{GM}\le\text{AM}$ inequality.

  • $n(1-\operatorname{AM}(1-\frac xn,1-\frac yn))=n(1-\frac 12(1-\frac xn+1-\frac yn))=\dfrac{x+y}2$

  • $n(1-\operatorname{HM}(1-\frac xn,1-\frac yn))=n(1-2(\frac 1{1-\frac xn}+\frac 1{1-\frac yn})^{-1})=\dfrac{nx+ny-2xy}{2n-x-y}\to \dfrac {x+y}2$

Since the expression is squeezed between the two, its limit is $\frac{x+y}2$ too.

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Since no one has responded to the final part of your question, $$ n \left( 1 - \sqrt{(1-x/n)(1-y/n)} \right) $$ is defined when the radical is nonnegative, that is, when $$ (1-x/n)(1-y/n) \geq 0 $$ and when $n \neq 0$ (since otherwise, the recipe given invokes division by zero). Since we may assume $n \neq 0$ when the sequence is defined, we multiply both sides of the inequality by $n^2$, which is positive so does not reverse the sense of the inequality. $$ (n - x)(n - y) \geq 0 $$ This is satisfied two ways

  • $n \geq x$ and $n \geq y$, so we have a product of nonnegatives on the left, or
  • $n \leq x$ and $n \leq y$, so we have a product of nonpositives on the left.

The first reduces to $n \geq \max\{x,y\}$ and the second to $n \leq \min\{x,y\}$. We can write this using only one expression: $$ \text{not } \min\{x,y\} < n < \max\{x,y\} \text{.} $$

However, for the purpose of obtaining a sequence as $n \rightarrow \infty$ for which every term of the sequence is defined, you want $n \geq \max\{x,y, 1\}$, so you also skip $n = 0$.

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  • $\begingroup$ thank you so so much for you detailed answer. That helped me incredibly! :) $\endgroup$ – Parinn Apr 20 at 2:45

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