# Showing $\lim_{m\rightarrow\infty} \left(1-\frac1{m^2}\right)^m=1$

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.

• Hint: With $m$ an integer, use binomial theorem and then estimate the terms of if $(1-\frac{1}{m^2})^m$. – Thomas Andrews Nov 20 '12 at 19:18
• 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. – Thomas Andrews Nov 20 '12 at 19:20
• @ThomasAndrews oh I've meant $1-\frac1m$ on the right side. – user50120 Nov 20 '12 at 19:24

When you get stuck, try proving something harder instead. Perhaps

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

• 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 $\frac{9}{16}\ge 1$. – hmakholm left over Monica Nov 20 '12 at 19:19
• @HenningMakholm Yeah, he corrected to it $\geq 1-\frac{1}m$ – Thomas Andrews Nov 20 '12 at 19:31

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

• 'reading the post' is your friend ;) although this one is the easiest solution... – user50120 Nov 20 '12 at 19:38

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.

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 $$y=-\frac{1}{m^2}$$

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 $$0$$,

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