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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.

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  • $\begingroup$ Hint: With $m$ an integer, use binomial theorem and then estimate the terms of if $(1-\frac{1}{m^2})^m$. $\endgroup$ – Thomas Andrews Nov 20 '12 at 19:18
  • $\begingroup$ I don't see how $$\left(1-\frac1{m^2}\right)^m\geq1$$ is ever true for $m$ a positive real number. Try $m=2$ and it doesn't work. $\endgroup$ – Thomas Andrews Nov 20 '12 at 19:20
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    $\begingroup$ @ThomasAndrews oh I've meant $1-\frac1m$ on the right side. $\endgroup$ – user50120 Nov 20 '12 at 19:24
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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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    $\begingroup$ On the other hand, I'm not sure I understand the question, because $(1-\frac 1{m^2})^m\ge 1$ (which the OP "knows by the Bernoulli-Inequality") is definitely not true in the arithmetic I'm used to. For $m=1$ I get $0\ge 1$ and for $m=2$ I get $\frac{9}{16}\ge 1$. $\endgroup$ – hmakholm left over Monica Nov 20 '12 at 19:19
  • $\begingroup$ @HenningMakholm Yeah, he corrected to it $\geq 1-\frac{1}m$ $\endgroup$ – Thomas Andrews Nov 20 '12 at 19:31
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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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    $\begingroup$ 'reading the post' is your friend ;) although this one is the easiest solution... $\endgroup$ – user50120 Nov 20 '12 at 19:38
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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.

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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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Since taking $\lim_{m \to \infty}\left( 1 - \frac{1}{m^{2}} \right)^{m}$ gives us the indeterminate form $1^{\infty}$, and, as the problem does not forbid us from using L'H Rule, let's use it here.

First, observe that $$ \ln \left( 1 - \frac{1}{m^{2}} \right)^{m} = m \ln \left( 1 - \frac{1}{m^{2}} \right) = \frac{\ln \left( 1 - \frac{1}{m^{2}} \right)}{\frac{1}{m}} $$

Hence, since the limit as $m \to \infty$ of the RHS of the above equality equals $0$,

$$ \lim_{m \to \infty}\left( 1 - \frac{1}{m^{2}} \right)^{m} = e^{\lim_{m \to \infty}\left( 1 - \frac{1}{m^{2}} \right)^{m}} = e^{0} = 1. $$

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