Weierstrass's approximation theorem with polynomials $p_n$ of degree $n$ for all $n.$ By Weierstrass's approximation theorem, every continuous function $f$ supported on an compact interval can be uniform approximated by polynomials. But, is it true that for every continuous $f$ on $[0,1]$, there exist a sequence of polynomials $\{f_n\}_{n=0}^{\infty}$ with $f_N=\sum_{n=0}^N a_n x^n$ ($a_N \neq 0$) such that $f_n$ converges to $f$? 
My attempt:  
For function $f=0$ , we can let $g_n=\frac{e^x}{n}$ and let $$f_N=\frac{\sum_0^N \frac{1}{n!}x^n}{N}$$ then we can find $f_n$ converges to $0$ uniformly. For each continuous function $g$ supported on $[0,1]$, if we can find a sequence of polynomials $\{h_n \}_{n=0}^{\infty}$ such that $h_n$ converges to $g$ uniformly and the degree of  $g_n$ is less than $n$, then with $f_N$ defined above, let $g_n=h_n+f_n$. We can prove that $g_n$ converges to $g$ uniformly.  
So, to prove the assertion above, it suffice to prove that following statement :
For every continuous function $f$ supported on $[0,1]$, we can find a sequence of polynomials $\{f_n \}_{n=0}^{\infty}$ with the degree of $f_n$ less than $n+1$, such that $f_n$ converges to $f$.  
Let $\{g_n \}$ be polynomials converges to $f$ uniformly, and let $k(n)$ denote the degree of $g_n$. Then we can contrust $f_n$ such that $f_0=f_1=...=f_{k(0)-1}=0$, $f_{k(0)}=g_0$. For $n={1,2,3,...} $ , if $k(n)\le n$ then $f_{n}=g_{n}$. If $k(n) \gt n$ , let $f_n=f_{n+1}=...=f_{k(n)-1}=f_{n-1}$ and $f_{k(n)}=g_n$. Since $f_n$ conveges to $f$ uniformly, the proof is complete. Is my proof correct?
 A: I think you have the right ideas, but I'm unsure about some of what you're doing. It may be just fine, but the notation is a little scrambled. 
Here's what I did. Suppose f is continuous on $[a,b].$ Then by Weierstrass, there is a sequence of polynomials $p_n\to f$ uniformly on $[a,b].$ Suppose the degrees of the $p_n$ are unbounded. Then we can choose a subsequence $q_n$ of $p_n$ such that the degrees of the $q_n$ strictly increase to $\infty.$ Clearly $q_n\to f$ uniformly on $[a,b].$ Now add terms of the form $x^k/k!,$ as you were doing, to get a sequence of the desired form. Suppose for example $\text { deg } q_1=4, \text { deg } q_2=8,$ $ \text { deg } q_3=9, \text { deg } q_4=12,\dots$ Then we construct a new sequence as below:
$$1,x,x^2/2!,x^3/3!, q_1(x), q_1(x)+x^5/5!, q_1(x)+x^6/6!, q_1(x)+x^7/7!,$$ $$ q_2(x), q_3(x), q_3(x)+x^{10}/10!, q_3(x)+x^{11}/11!, q_4(x), \dots $$
This sequence does what you want.
If the degress of the $p_n$ are bounded, say by $d,$ then we can define
$$q_n(x) = p_n(x) + x^{d+n}/(d+n)!.$$
Then $\deg q_n = d+n$ and $q_n\to f$ uniformly on $[a,b].$ Since the degrees increase to $\infty,$ the above applies to $q_n$ and again we get the result.
