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I want to prove $\lim_{m\rightarrow\infty} \left(1-\frac1{m^2}\right)^m=1$ without using the fact that $\lim_{m\rightarrow\infty}\left(1+\frac1m\right)^m=\mathrm e$.

I know by the Bernoulli-Inequality $\left(1-\frac1{m^2}\right)^m\geq1-\frac1m$ But now I don't know how to show $\left(1-\frac1{m^2}\right)^m\leq1$ for all $m\in\mathbb N$.

Anybody could help? Thanks.

  • 1
    @ThomasAndrews oh I've meant $1-\frac1m$ on the right side.2012-11-20

5 Answers 5

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I have proof on this site: $e^x=\displaystyle\lim_{n \to \infty}(1+\frac{x}{n})^n$

-here I am gonna modificate some elements and others not explaning .

So, we gonna need this binomial expansion:

$(1-y)^n=\binom{n}{0}y^0-\binom{n}{1}y^1+\binom{n}{2}y^2+...+(-1)^{n-1}\binom{n}{n-1}y^{n-1}+(-1)^{n}\binom{n}{n}y^n$

Hense:

$(1-y)^n=1-n*y+\frac{(n-1)n}{2!}y^2+...+(-1)^{n-1}*n*y^{n-1}+(-1)^{n}*y^n$

Here: $n=m$ and $y=\frac{1}{m^2}$ :

$(1-\frac{1}{m^2})^m=1-m*\frac{1}{m^2}+\frac{(m-1)m}{2!}(\frac{1}{m^2})^2+...+(-1)^{m-1}*m*(\frac{1}{m^2})^{m-1}+(-1)^{m}*(\frac{1}{m^2})^m$

$=1-\frac{1}{m}+\frac{1}{2}\frac{m-1}{m^3}+...=1-\frac{1}{m}+\frac{1}{2}(\frac{1}{m^2}-\frac{1}{m^3})+...$

Is not hard to see here that we have 1 + something (very) small: $\displaystyle\lim_{n \to \infty}(1-\frac{1}{m^2})^m=1+0+0+...=1$ Q.E.D.

PS: Alternatively we could use:

$(1+y)^n=\binom{n}{0}y^0+\binom{n}{1}y^1+\binom{n}{2}y^2+...+\binom{n}{n-1}y^{n-1}+\binom{n}{n}y^n$ ; where $y=-\frac{1}{m^2}$

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$\left(1-\frac{1}{m^2}\right)^m=\left(1-\frac{1}{m}\right)^m\times \,\,\left(1+\frac{1}{m}\right)^m$

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    'reading the post' is your friend ;) although this one is the easiest solution...2012-11-20
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When you get stuck, try proving something harder instead. Perhaps

$\left( 1-\frac1{a}\right)^n \le 1 $ for all $a\ge 1$ and $n\ge 1$?

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    @HenningMakholm Yeah, he corrected to it $\geq 1-\frac{1}m$2012-11-20
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Apply $\ln$, rewrite with $1/m$ in the denominator and then use L'Hospital's rule to bring the expression into the form of a rational function.