I want to prove that if we have an exponential distribution with parameter $\lambda$, we have that $P(X \le x)=\lambda x + o(x)$. I want to do this by using Taylor-series and the lagrange remainder theorem. There is one detail that I am not able to complete, basically, I can only show that this holds for an interval $[0,a]$ for any $a$, but not $[0, \infty)$.
Here are my details.
I need to show that $1-e^{-\lambda x}=\lambda x+o(x)$
The n taylor polynomial on any interval [0,a] becomes:
$\Sigma_{1}^n[\frac{(-1)^{n+1}\lambda^n*x^n}{n!}]$ . The remainder term is $\frac{(-1)^{n+2}\lambda^{n+1}*e^{-\lambda c}x^{n+1}}{(n+1)!}$, for a c in the interval. An upper bound for this is: $\frac{\lambda^{n+1}*a^{n+1}}{(n+1)!}$, and this goes to zero in our interval. I have two problems now:
1.I almost have my result, but the problem is that n is not bounded, if the sum was finite I see from $\Sigma_{1}^n[\frac{(-1)^{n+1}\lambda^n*x^n}{n!}]$ that I would get $\lambda x + o(x)$. But if the sum was infintie how can I then be sure that last parts of the sum will be o(x)?, what I mean is, if you add a finite number of variables whose limit is 0, then this limit is 0, but this does not hold for an infinite sum?
2. I only proved in on an interval [0,a], how do I extend it to $[0,\infty)$?
The derivative of the function $u$ defined by $u(x)=\mathrm e^{-\lambda x}$ is such that $u'(x)=-\lambda \mathrm e^{-\lambda x}$. By Taylor formula, for every nonnegative $x$, there exists some $z(x)$ in $[0,x]$ such that $u(0)-u(x)=-xu'(z(x))$, that is, $1-\mathrm e^{-\lambda x}=\lambda x\mathrm e^{-\lambda z(x)}$. When $x\to0$, $z(x)\to0$ hence $\mathrm e^{-\lambda z(x)}\to1$. This proves the result.
Sorry but I fail to understand this question.