Is $(1+m)^{1/m} < (1+\frac{1}{m})^m$?

Question

If $(1+\frac{1}{m})^m<e$; $m \in N $

How can we prove $(1+m)<e^m$ ?

I've been able to prove that above statement by mathematical induction but I'm unable to see it as a direct consequence of the give statement above.

If I can prove this $(1+m)^{1/m} < (1+\frac{1}{m})^m$ without using mathematical induction I shall be able to rest my case. Please help

Answer 1

By Bernoulli inequality

$$(1+m)^{1/m}\le1+\frac1m m=2<e$$

and

$$\left(1+\frac{1}{m}\right)^m\ge 1+m\frac 1m=2$$

As an alternative by AM-GM inequality

$$(1+m)^{1/m}=\sqrt[m]{\overbrace{1\cdot 1\cdot \ldots \cdot(1+m)}^{m\, terms}}\le \frac{\overbrace{1+ 1+ \ldots +(1+m)}^{m\, terms}}{m}=\frac{2m}m=2$$

Answer 2

Well, actually $(1+\frac1m)^{m}<e$ holds for all $m\geq 0$. Thus in particular for $m'=\frac1m$ the inequality holds. By substitution we obtain $$(1+m')^{1/m'}<e$$ from which you conclude using that functions of the type $x^q$ are increasing for all $q\geq0$.

Answer 3

Alternatively note that $$ e^m = \sum_{j=0}^\infty \frac{m^j}{j!} = 1 + m + \sum_{j=2}^\infty \frac{m^j}{j!} > 1+m, $$ for $m > 0$.

Answer 4

Recall Bernoulli's inequality:

$$(1+x)^n \geq 1+nx, \forall x>-1.$$

Thus,we have $$1+m=\left(1+m^2 \cdot \frac{1}{m}\right)\leq \left(1+\frac{1}{m}\right)^{m^2}=\left[\left(1+\frac{1}{m}\right)^{m}\right]^m\leq e^m. $$