Vtyjkyuk

If $(1+frac1m)^m<e$; $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 $(1+m)^1/m < (1+frac1m)^m$ without using mathematical induction I shall be able to rest my case. Please help

By Bernoulli inequality

$$(1+m)^1/mle1+frac1m m=2<e$$

and

$$left(1+frac1mright)^mge 1+mfrac 1m=2$$

As an alternative by AM-GM inequality

$$(1+m)^1/m=sqrt[m]overbrace1cdot 1cdot ldots cdot(1+m)^m, termsle fracoverbrace1+ 1+ ldots +(1+m)^m, termsm=frac2mm=2$$

Well, actually $(1+frac1m)^m<e$ holds for all $mgeq 0$. 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 $qgeq0$.

Alternatively note that
$$ e^m = sum_j=0^infty fracm^jj! = 1 + m + sum_j=2^infty fracm^jj! > 1+m, $$
for $m > 0$.

Recall Bernoulli's inequality:

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

Thus,we have $$1+m=left(1+m^2 cdot frac1mright)leq left(1+frac1mright)^m^2=left[left(1+frac1mright)^mright]^mleq e^m. $$