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I am trying to Prove Uniform convergence of the series of functions :${(1+ (x/n))}^n$ in Real numbers (and the difference between the real numbers and [0,1]), I got to $e^x$ but can not prove the Uniform convergence.

I would really appreciate some help, thanks in advance.

razivo
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Lahav222
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  • You could use Dini's theorem. – nejimban Sep 14 '21 at 19:15
  • Yes, I agree with @nejimban, we often do not bother proving uniform convergence because dominated convergence theorem is used instead, but you can use Dini as shown in this other post https://math.stackexchange.com/q/3175585/399263 (though with $e^{-x}$ but it's similar). – zwim Sep 14 '21 at 19:23

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$f_n(x)=(1+x.n)^n$ does NOT converge uniformly on $\Bbb R$ to $e^x.$ For $n\in\Bbb N$ we have $\sup_{x\in\Bbb R}|e^x-f_n(x)|\ge |e^n-f_n(n)|=e^n-2^n>1/2.$

For $y\in [0,1]$ we have $$\frac {y}{1+y}\le \ln (1+y)\le y$$ which can be applied with $y=x/n$ to show that $f_n$ converges uniformly on $[0,1]$ to $\exp.$

Let $y\in [0,1].$ Then $$\ln (1+y)=\int_1^{1+y}\frac 1 t dt \le \int_1^{1+y}1\cdot dt=y.$$ Let $1+y=(1-z)^{-1}.$ Then $z=\frac {y}{1+y}\in [0,1/2].$ We have $$\ln (1+y)=|\ln (1-z)|=\int_{1-z}^1\frac 1 t dt\ge \int_{1-z}^1 1\cdot dt=z=\frac {y}{1+y}.$$

DanielWainfleet
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