Sequence of functions converges normally to $e^z$ Consider the functions $f_n(z)=(1+\frac{z}{n})^n$ for each $n=1,2,3,...$ Use the fact that $f_n(x)$ $\rightarrow e^x $ for each $x \in \Bbb R $ to prove that $f_n(z) \rightarrow e^z$ normally in $\Bbb C$. 

My work: Suppose $f_n(z)=(1+\frac{z}{n})^n$ for each $n=1,2,3,...$ Since $1+x \le e^x$ for $x \ge 0$, it follows that $|f_n(z)| \le e^{|z|}$, for all $z$. So, ${f_n}$ is a normal family. And $f_n(x)$ $\rightarrow e^x $ for each $x \ge 0$. Did I miss anything? Now, How to use Vitali's theorem here to prove $f_n(z) \rightarrow e^z$ uniformly on compact sets. Your kind help will be appreciated. Thank you so much!
I have no idea how you deduced that your family is normal. But in order to that $(f_n)_{n\in\mathbb N}$ converge normally to the exponential function, you use the fact that, for each $z\in\mathbb C$ and each natural $n$, if you expand $\left(1+\frac zn\right)^n$, then you get a polynomial $1+z+a_{n,2}z^2+a_{n,3}z^3+\cdots+a_{n,n}z^n$ with $a_{k,n}\leqslant\frac1{k!}$ and therefore\begin{align}\left|e^z-\left(1+\frac zn\right)^n\right|&amp;=\left|\left(1+z+\frac{z^2}{2!}+\frac{z^3}{3!}+\cdots\right)-\left(1+z+a_{n,2}z^2+a_{n,3}z^3+\cdots+a_{n,n}z^n\right)\right|\\&amp;=\left|\left(\frac1{2!}-a_{2,k}\right)z^2+\left(\frac1{3!}-a_{3,k}\right)z^3+\cdots+\left(\frac1{n!}-a_{n,k}\right)z^n+\frac{z^{n+1}}{(n+1)!}+\cdots\right|\\&amp;\leqslant e^{|z|}-\left(1+\frac{|z|}{n!}\right)^n.\end{align}So, if $K$ is compact and $M=\max_{z\in K}|z|$, you can use the fact that$$(\forall z\in K)(\forall n\in\mathbb{N}):\left|e^z-\left(1+\frac zn\right)^n\right|\leqslant e^M-\left(1+\frac Mn\right)^n.$$
Let f and g {/belongs to }R[a,b] .Let p be a partition such that P=[X0,X1],.......[Xn-1,Xn]. Suppose X0=a and Xn=b.let f
and g continuous and f≥g then g(x)-F(x)≤0 so g(x)-f(x) belongs to R[a, b] then {${/int_a^b(g(x)-f(x)}=/Int_a^X1(g(x)-f(x))+Int_X1^X2(g(x)-f(x))...........Int_Xn-1^b(g(x)) -f(x)) ≤0*(b-a) so Int_a^X1(g) +.......-(Int_a^X1(f)) ≤0 so Int _a^b(g) -Int_a^b(f) ≤0$