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Let

$$ f_{n}: \mathbb{R} \xrightarrow{} \mathbb{R}, x \mapsto f_{n}(x) = \left( 1 + \frac{x}{n}\right)^{n} $$

Now we know that for all $x \in \mathbb{R}$ we have

$$ \lim_\limits{n \to \infty} f_{n}(x) = \lim_\limits{n \to \infty} \left( 1 + \frac{x}{n}\right)^{n} = \exp(x) $$ pointwise. My question is now, whether the sequence $\{f_{n}\}_{n \in \mathbb{N}}$ converges uniformly to $\exp(x)$

That $f_{n}$ converges uniformly on any compact interval follows immediatley from Dini's Theorem. Does it also converges uniformly on any finite subinterval of $\mathbb{R}$?

  • You are asking for uniform convergence on the entire real line, right? – Sarvesh Ravichandran Iyer Sep 12 '18 at 14:48
  • exactly! Maybe the uniform convergence is only satisfied on finite intervals? –  Sep 12 '18 at 14:52
  • On compacts sure, it does. But if you want convergence on the whole real line, I’m quite sure there isn’t. – tommy1996q Sep 12 '18 at 14:52
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    $f_n(n)\to \infty$ as $n\to \infty$. – Riemann Sep 12 '18 at 14:54
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    See also https://math.stackexchange.com/questions/1108581/uniform-convergence-to-the-exponential-function-over-a-compact-interval and https://math.stackexchange.com/questions/1744506/proving-uniform-convergence-of-1-fracxnn-to-ex-on-compact-intervals – Robert Z Sep 12 '18 at 14:57
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    @Riemann Wait, that shouldn’t be a problem. The issue is that the sup of the difference btw exponential and that function doesn’t go to zero for $n \to \infty$ if you take $x \in \mathbb{R}$, or am I making some trivial mistake? – tommy1996q Sep 12 '18 at 14:57
  • but why does it not converge uniformly on any finite open interval? –  Sep 12 '18 at 15:04

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On the whole real line, no. That’s because for every $n$ you can choose $x$ accordingly so that the difference btw exponential and $(1+x/n)^n$ (basically a polynomial of degree $n$) becomes arbitrarely large.

tommy1996q
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