I have a positive sequence which converges to zero, i.e. $a_k \geq 0 \;, \forall k \in \mathbb{N}$ and $\lim_{k\rightarrow \infty} a_k = 0$. Does there exist another sequence $b_k$ with the property that for every $k>0$ it holds that $a_{k+p} < b_{k} \;, \forall p > 0$ and in addition, $\lim_{k\rightarrow \infty} b_k = 0$?

In other words, $b_{k}$ is an upper bound for $a_{k+p}\;, p>0$ converging to $0$.

Intuitively i would say yes and argue: Since $a_k$ converges to $0$, we have that, for all $k>0$ there exists an $\epsilon(k) > 0$ such that $a_{k+p} < \epsilon(k) \;, \forall p>0$. Now we construct a seuqence $b_k$ such that for all $p>0$ we have $a_{k+p} < b_k < \text{max}(a_{k+1}, a_{k+2}, \dots) + \frac{1}{k}$. Then, by taking the limit of all sequences in the last inequality, we have $0 \leq \lim_{k\rightarrow \infty} b_k \leq 0$ which concludes the proof.

I am not sure if this is correct. Would be very nice if someone gives a little feedback.



Now we construct a sequence $b_k$ such that for all $k>0$ we have $a_{k+p}<b_k<\max(a_{k+1},a_{k+2},\ldots)+\frac{1}{k}$...

Did you perhaps mean "for all $p>0$" here?

The argument itself seems sound. You can in fact even set $$ b_k = \max(a_{k+1},a_{k+2},\ldots) + \frac{1}{k}. $$ It's worth noting (if you haven't already) that you've in fact used the limit superior here. Note that by the equality $$ \limsup_{k\to\infty} a_k = \lim_{k\to\infty}\sup\{a_k, a_{k+1},\ldots\} $$ if you take the limit as $k\to\infty$ in the inequality $$ a_k < b_k = \max(a_{k+1},a_{k+2},\ldots) + \frac{1}{k} $$ you get $$ 0 \leq \lim_{k\to\infty} b_k \leq \limsup_{k\to\infty} a_k + \lim_{k\to\infty} \frac{1}{k} = 0 + 0. $$

  • $\begingroup$ Thank you very much for the feedback. Yes, I indeed meant "for all $p>0$". I will correct it. $\endgroup$ – Rufio Jul 8 '15 at 7:11

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