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In this post a user made the following claim:

Claim: Suppose $\rho_n$ is a sequence of real non-negative numbers converging to $0$. Suppose $x\in (0,1)$. Then $$ \sum_{N+1}^\infty (n+1)(1-x)^2x^n\rho_n $$ is small for $N$ large, and $N$ can be chosen independently from $x$.

I'm not seeing how this is true. I calculate that $$ (1-x)^2\sum_{N+1}^\infty (n+1)x^n = x^{n+1}(n+2 - (n+1)x) $$ so this part does not seem small independently of $x$. Can anyone justify this claim, or else confirm it is not true? (Perhaps, but I don't think so, there is some other contextual detail which makes his claim true in the linked answer.)

Community
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Eric Auld
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1 Answers1

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Let $\mathop {\lim }\limits_{n \to \infty } {\rho _n} = 0$, then $\forall \varepsilon > 0,\exists {N_\varepsilon } \in \mathbb{N},n \geqslant {N_\varepsilon } \Rightarrow {\rho _n} < \varepsilon $

So,

$\sum\limits_{n = {N_\varepsilon } + 1}^\infty {\left( {n + 1} \right){{\left( {1 - x} \right)}^2}{x^n}{\rho _n}} \leqslant \varepsilon \sum\limits_{n = {N_\varepsilon } + 1}^\infty {\left( {n + 1} \right){{\left( {1 - x} \right)}^2}{x^n}} = \varepsilon {\left( {1 - x} \right)^2}\left( {n + 2 - \left( {n + 1} \right)x} \right)$

which is arbitrarily small, independent of $x \in \left( {0,1} \right)$ (it is only dependent on the sequence ${\rho _n}$ and ${N_\varepsilon }$ is dependent only on $\varepsilon $).

Alen
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