When calculating the limit $L=\big( 1 - \frac 1 {\sqrt n}\big)^n$, $n\to\infty$, what allows me to do the following:

$$ L=\lim \left(\left(1-\frac {1}{\sqrt n}\right)^\sqrt{n} \right)^\sqrt{n} $$

As the term inside the outer parenthesis goes to $e^{-1}$, we have $L=\lim e^{-\sqrt n}=0$.

It's like we're distributing the parenthesis somehow:

$$\lim \left(\left(1-\frac {1}{\sqrt n}\right)^\sqrt{n} \right)^\sqrt{n}=\lim \left(\lim\left (1-\frac {1}{\sqrt n}\right)^\sqrt{n} \right)^\sqrt{n}=\lim e^{-\sqrt n}$$

The question is: Why can we do this? Which property are we using?

  • $\begingroup$ What is the question? $\endgroup$ – Jan Mar 3 '16 at 17:59
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    $\begingroup$ If $n$ is large enough, then the inner power is close to $e^{-1}$, and in particular is less than $1/2$. And $(1/2)^{\sqrt{n}}\to 0$. $\endgroup$ – André Nicolas Mar 3 '16 at 17:59
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    $\begingroup$ @AndréNicolas Yes, I understand that, but I'm looking for a more rigorous proof of this property (and a generalization, if possible). $\endgroup$ – limittroubled Mar 3 '16 at 18:00
  • $\begingroup$ @JanEerland: OP is asking what justifies breaking up the expression into nested powers. $\endgroup$ – Brian Tung Mar 3 '16 at 18:01
  • $\begingroup$ @limittroubled: If you accept the nesting uncritically without the limit, then just use the definition of the limit as $n \to \infty$. First show that the inner power is less than $1/2$. Then, let $N = -\lg \varepsilon$, and show that for any $n > N$, the expression is less than $\varepsilon$. $\endgroup$ – Brian Tung Mar 3 '16 at 18:04

You are looking for the following statement.

Let $(a_n)$ be a converging sequence of non-negative real numbers with $\lvert \lim_{n → ∞} a_n \rvert < 1$. Furthermore, let $(e_n)$ be an unbounded, increasing sequence. Then $({a_n}^{e_n})$ converges to zero.

Proof idea. Let $a = \lim_{n → ∞} a_n$. Let $K > 1$ be a real number. Then $({a_n}^K)$ converges to $a^K$ by limit theorems. Since $e_n > K$ for large enough $n$, almost all members of $({a_n}^{e_n})$ are smaller than $a^K$, so both limes inferior and limes superior of the sequence $({a_n}^{e_n})$ lie within the interval $[0..a^K]$.

Because $K$ was arbitrary and $\lvert a \rvert < 1$, the sequence must converge to zero.


One may use, as $x \to 0$, the classic Taylor expansion: $$ \ln (1+x)=x-\frac{x^2}2+O(x^3) $$ giving, as $n \to \infty$, $$ n\ln \left(1-\frac1{\sqrt{n}}\right)=n \times\left(-\frac1{\sqrt{n}}-\frac1{2n}+O\left(\frac1{n^{3/2}}\right)\right)=-\sqrt{n}-\frac12+O\left(\frac1{\sqrt{n}}\right) $$ then, as $n \to \infty$, $$ \left(1-\frac {1}{\sqrt n}\right)^n=e^{-\sqrt{n}-1/2}\left(1+O\left(\frac1{\sqrt{n}}\right)\right) \to 0. $$


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