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Let $X = P[0,1]$ be the space of polynomials on the interval $[0,1]$ with infinity norm, i.e. $||f(x)||_{\infty} = max_{[0,1]} |f(x)|$.

Suppose I have the following sequence in $X$: $\{f_n(x)\}_{n=1}^{\infty}$ where $f_n(x) = 1 + x + \frac{x^2}{2} + \frac{x^3}{3!} + ... + \frac{x^n}{n!}$. How can I show that this sequence is Cauchy?

Here is my attempt:

Fix some small value $\epsilon$, then I must show that there exists some natural number $N$ such that for all $m, n \geq N$, $||f_n(x) - f_m(x)|| \leq \epsilon$.

I begin with $||f_n(x) - f_m(x)|| = max_{[0,1]}|f_n(x) - f_m(x)|$

But I hit a dead end because I don't know if I can use any information about this sequence from an outside perspective (like that it converges pointwise to $e^x$). How can I proceed?

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  • $\begingroup$ Prove it converges uniformly? There's a famous test for uniform convergence of series. $\endgroup$ – Lord Shark the Unknown May 12 '17 at 15:31
  • $\begingroup$ But doesn't it converge pointwise? $\endgroup$ – Nażysław Zbyłutowicz May 12 '17 at 15:36
  • $\begingroup$ It certainly converges pointwise; does it converge uniformly? $\endgroup$ – Lord Shark the Unknown May 12 '17 at 15:37
  • $\begingroup$ It doesn't, the rate of convergence is different for each $x \in [0,1]$ as far as I can tell. $\endgroup$ – Nażysław Zbyłutowicz May 12 '17 at 15:39
  • $\begingroup$ If it isn't uniformly convergent then your attempts to prove convergence in the infinity norm are doomed! $\endgroup$ – Lord Shark the Unknown May 12 '17 at 15:48
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Hint. The sequence $(f_n)_n$ converges pointwise to $e^x$. Show that for $0\leq k\leq n$, $g_n^{(k)}=g_{n-k}$ is increasing and non-negative in $[0,1]$ where $g_n(x)=e^x-f_n(x)$. Hence $$\max_{[0,1]}|e^x-f_n(x)|=|e-f_n(1)|$$ which goes to zero as $n\to \infty$. Finally note that $$\max_{[0,1]}|f_n(x) - f_m(x)|\leq \max_{[0,1]}|e^x-f_n(x)|+\max_{[0,1]}|e^x-f_m(x)|.$$

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Alternative hint: if a power series $\sum_{n=0}^\infty a_nx^n$ converges for $|x|<R$, then for any $\epsilon>0$ it converges uniformly for $|x|\le R-\epsilon$. You can prove this either directly or using the Weierstrass M-test.

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