I have a cumulative probability distribution function F(x).
I need to prove that
$\sum_{n=2}^\infty n(F(x)^{n-1} - F(x)^n)$ does not converge for all $x \in \mathbb{R^+} - \{x:F(x)\ne 0\}$
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1It is visibly not pointwise convergence (for an $F$ that is $0$ in $0$, it suffices to take $x=0$ to get a convergent series). Thus convergence should be as a series of functions... – Jean Marie Jan 27 '18 at 06:53
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You can factor $(1-F(x))$ in your series... It will look simpler. But after that ... – Jean Marie Jan 27 '18 at 06:56
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My apologies, ....only for $x$ such that $F(x) \ne 0$ – Andy Jan 27 '18 at 07:10
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That's wrong. $F(x)\in[0,1]$, and $\displaystyle\sum_{n=2}^\infty n(y^{n-1} - y^n)$ is convergent for all $y\in[0,1]$. – Jan 27 '18 at 07:20
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@ProfessorVector could you explain why ? – Andy Jan 27 '18 at 08:37
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That's wrong. For $F(x)<1$:$$\sum_{n=2}^\infty n(F(x)^{n-1} - F(x)^n)=(1-F(x))\sum_{n=2}^\infty nF(x)^{n-1}\\=(1-F(x))\dfrac{2F(x)-F^2(x)}{(1-F(x))^2}\\=F(x)\dfrac{2-F(x)}{1-F(x)}$$and for $F(x)=1$:$$\sum_{n=2}^\infty n(F(x)^{n-1} - F(x)^n)=0$$which is convergent at both cases.
Mostafa Ayaz
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This is wrong, too. $$\sum_{n=2}^\infty nF(x)^{n-1}=\sum_{n=1}^\infty nF(x)^{n-1}-1=\frac1{(1-F(x))^2}-1.$$ – Jan 27 '18 at 08:50
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I don't agree at all with you when you say: $F(x)<1$ for $x \in \mathbb{R}$ ? If, for instance, $F$ is the "ramp function" associated to a Uniform on [0,1], you have $F(x)=1$ for all $x \in (1,\infty)$ ... – Jean Marie Jan 27 '18 at 10:17
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