Prove that if $f$ is defined for $|x|< r$ and if there exists a constant $B$ such that
$$| f^n(x) |\le B$$
for all $|x|< r$ and $n \in \mathbb N$, then the Taylor series expansion :
$$\sum_{n=0}^\infty \frac{f^{(n)}(0)x^n}{n!}$$
converges to $f(x)$ for $|x|< r$.
What I did : I know a little bit about Remainder approximation theorem for Taylor series , So we're given $|f^{(n)}(x)| \le B$, thus $|f^{(n+1)}(x)|\le B$, this implies :
$$| R_n(x) | \le B\frac{x^{n+1}}{(n+1)!},$$
so by remainder approximation theorem, the Taylor series must converge to $f(x)$.
Is this correct?
Best Answer
A Taylor series converges if and only if its remainder term, $R_n(x)$ tends to zero. This is just a consequence from taylors theorem
$\mathbf{Theorem:}$ If $f(x)=T_n(x)+R_n(x)$ , where $T_n$ is the nth degree Taylor polynomial of f at a and $\lim_{n \to \infty} R_n(x)=0$ for $|x-a| < R$ , then f is equal to the sum of its taylor series on the interval $|x-a| < R$ That is because, as you suggested, we have $$|R_n(x)| \le B \frac{x^{n+1}}{(n+1)!}$$
and
$\mathbf{Taylors}$ $\mathbf{Inequality:}$ If $| f^{n+1}(x) | \le M$ for $|x-a| \le d$ , then the remainder, $R_n (x)$ of the Taylor series satisfies the inequality $| R_n (x) | \le \frac{M}{(n+1)!}|x-a|^{n+1}$ for $| x-a | \le d$
so now you must consider this,
$$\lim_{n \to \infty} B \frac{x^{n+1}}{(n+1)!}$$
$$= B \lim_{n \to \infty} \frac{x^{n+1}}{(n+1)!} $$
and we that $$B \lim_{n \to \infty} \frac{x^{n+1}}{(n+1)!} = B \bullet 0 = 0$$
thus we have that $|R_n(x)| \le 0$ i.e., the remainder will tend to zero as n approaches infinity , and the proof is finished.
Thus, we approach the function itself.