$F_n(x)=\left(1+ \frac xn \right)^n$

Bounded and uniformly bounded Sequence of functions

I tried as someone solved in this link but I think i should not take n =1 So I took n tend to infinite Then it became $e^x$ $$ \bbox[5px,border:2px solid red] { e^x=\lim_{n\to\infty} \left( 1+\frac{x}{n} \right)^n \qquad (2) } $$

Now can I say that function is less then or equal to $e^x$ ? I'm just talking about boundedness not uniformly boundedness since I think it is not uniform bounded if I don't restrict x to closed interval.


1 Answer 1



Let $$g_n (x)=n\ln (1+\frac {x}{n})-x $$

$$g'_n (x)=\frac {n}{n+x}-1=\frac {-x}{n+x} $$

$g_n $ is decreasing at $[0,+\infty) $, then

$g_n (x)\le g_n(0 )=0$.

$$\implies n\ln(1+\frac {x}{n})\le x $$ $$\implies (1+\frac {x}{n})^n\le e^x $$

for $x\ge 0$.

  • $\begingroup$ Don't you think it's a bit complicated ..? If we just let n tend to infinity , what will be the problem? $\endgroup$ Jul 25, 2017 at 11:23
  • $\begingroup$ And also that should be n, not 1/n in the power don't you think? $\endgroup$ Jul 25, 2017 at 11:24
  • $\begingroup$ @Ikuyuki Yes thnx $\endgroup$ Jul 25, 2017 at 11:26

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .