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Let $\{u_n\}$ be a sequence of nonnegative numbers satisfying the condition $$ \tag{1} u_{n+1}\leq (1-\alpha_n)u_n+\beta_n \quad \forall n\in\mathbb{N}, $$ where $\{\alpha_n\}$ and $\{\beta_n\}$ are sequences of real numbers such that

$\tag{2}\displaystyle\lim_{n\rightarrow\infty}\alpha_n=0$

$\tag{3}\displaystyle\sum_{n=0}^{\infty}\alpha_n=\infty$

$\tag{4} \displaystyle\sum_{n=0}^{\infty}\beta_n<\infty$

Prove that $$\displaystyle\lim_{n\rightarrow\infty}u_n=0$$

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    Hint: Try to prove the contrapositive. "If u_n does not tend to 0, then there exists an epsilon and n_0 such that |u_n| >= epsilon for all n >= n_0..."2012-05-03
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    @AdamRubinson, that's not the contrapositive exactly, since "If u_n doesn't tend to 0" means what you wrote **or** that there is no limit at all, which would allow for the series to visit as close to 0 as you like infinitely often.2012-05-03
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    Actually davin, you are right. I guess what I meant to say was: "there exists an epsilon such that for every n_o, there is an n_1 > n_0 such that |u_(n_1)| > epsilon". But the main point is that, assuming the result in the OP is true, it should be provable using epsilon's and delta's (if OP is familiar with these), probably along with AOL.2012-05-03
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    drmath: Any luck with my answer below?2012-05-07

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