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*All sequences considered are non-negative.

Suppose we are given sequences $a_n,b_n$ such that $\sum a_n=1$ and $b_n$ is a sequence going to infinity. I am searching for any sequence $c_n$ with the properties that

(1) $c_n\to \infty$

(2) $c_n\le b_n$

(3) $c_n\cdot\max\lbrace a_k:b_n\le k\le 2b_n\rbrace\to 0$

Does such a sequence $c_n$ exists?

The main difficulty I am having is due to the fact that $na_n$ need not go to zero, so $b_n \max\lbrace a_k:b_n\le k\le 2b_n\rbrace$ need not go to zero (otherwise I would just take $c_n=b_n)$. I am not sure if it is true that there is a $c_n$ (e.g., $c_n= \log b_n$) that satisfies (3).

I have tried to consider $c_n=b_n^a$ with $0<a<1$, however I am not sure if (even for small $a$) we have $b_n^a\max\lbrace a_k:b_n\le k\le 2b_n\rbrace\to 0$. Perhaps the choice of $c_n$ should depend on $a_n$ as well.

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  • $\begingroup$ Just to make it more clear: are you looking for the three sequences, or a proof that given $\{a_n\}$ and $\{b_n\}$ then $\{c_n\}$ exists? $\endgroup$ – ajotatxe Jan 3 '17 at 10:42
  • $\begingroup$ The latter. Given $a_n, b_n$, I am seeking $c_n$. $\endgroup$ – The Substitute Jan 3 '17 at 10:43
  • $\begingroup$ @TheSubstitute But in the parraph below (3) you talk of $\;b_n\cdot \max...\;$ , so which one is it? $\endgroup$ – DonAntonio Jan 3 '17 at 10:47
  • $\begingroup$ @DonAntonio That paragraph below (3) is a fact that is causing me difficulty. If the limit $b_n∗max$ was zero, I would just take $c_n=b_n$. However, $b_n∗max$ need not be zero unless $a_n$ is a monotone decreasing sequence. $\endgroup$ – The Substitute Jan 3 '17 at 11:25
  • $\begingroup$ If $a_n>0$ and $\Sigma a_n=1$ then $na_n$ must to go zero just FYI $\endgroup$ – Arjang Jan 4 '17 at 12:41
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Define $$d_n=\max\{a_k:b_n\le k\le 2b_n\}$$

Let's show that $d_n\to 0$. Let $\epsilon>0$. There exists some $K$ such that $a_n<\epsilon$ whenever $n\ge K$. There exists also some $N$ such that $b_n>K$ whenever $n\ge N$. Therefore, for $n\ge N$ we have that $k\ge b_n>K$ for every integer $k\in[b_n,2b_n]$, and then $a_k<\epsilon$. Thus, $\max\{a_k:b_n\le k\le 2b_n\}<\epsilon$.

Now we know that $d_n\to 0$, so $d_n|\log d_n|\to 0$, and $|\log d_n|\to\infty$, so you can define $c_n=\min\{|\log d_n|,b_n\}$.

If I'm not wrong this sequence holds the three properties:

  1. $c_n\to\infty$ because is the minimum of two sequences that go to infinity.
  2. $c_n\le b_n$ is obvious.
  3. $0\le \lim c_nd_n\le\lim |\log d_n|d_n=0$

Remarks:

  • To avoid the $\max$ of an empty set, we need to assume that $b_n\ge1/2$ for every $n\in \Bbb N$.
  • The condition $\sum a_n=1$ seems too strong. Only the convergence to $0$ is needed.
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