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Let $(v_k)_{k\geq0}\subset \mathbb{R}_+$ be s sequence such that $v_{k+1}\leq u_kv_k$, where $(u_k)_{k\geq 0} \subset [0,1]$ satisfies $$\sum_{k=0}^{\infty}(1-u_k)=\infty$$ Show that $v_k\rightarrow 0$.

Any hints?

user112358
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    What if you try to show that the product of all $u_k$ (for $k$ going from one to $n$) converges to $0$ (with increasing $n$)? – arch1t3cht30 Apr 18 '17 at 20:45
  • Also, on MathStackexchange, you should generally always show your efforts on the question, so please explain what you've tried already. For example, I can't know whether my previous comment will be of any worth to you because I don't know whether you have already considered that. – arch1t3cht30 Apr 18 '17 at 20:47

1 Answers1

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Let $N \in \Bbb N$. For $n \ge N$,

$$\prod_{k=0}^{n-1}u_k \le^{AM-GM} \left(\frac{\sum_{k=0}^{n-1} u_k}{n}\right)^n = \left( 1 - \frac{\sum_{k=0}^{n-1} (1-u_k)}{n}\right)^n \le \left(1 - \frac{\sum_{k=0}^{N-1}(1-u_k)}{n}\right)^n$$

Taking $n \to \infty$,

$$\lim_n \prod_{k=0}^{n-1}u_k \le \exp\left( -\sum_{k=0}^{N-1} (1-u_k)\right)$$

This is true for all $N \in \Bbb N$. Taking $N \to \infty$, we get:

$$\lim_n \prod_{k=0}^{n-1}u_k = 0$$

Now $v_n \le u_{n-1}v_{n-1} \le u_{n-1}u_{n-2}v_{n-2} \le \cdots \le v_0\prod_{k=0}^{n-1} u_k$. This shows $v_n \to 0$.