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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.

The Substitute
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  • 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? – ajotatxe Jan 03 '17 at 10:42
  • The latter. Given $a_n, b_n$, I am seeking $c_n$. – The Substitute Jan 03 '17 at 10:43
  • @TheSubstitute But in the parraph below (3) you talk of $;b_n\cdot \max...;$ , so which one is it? – DonAntonio Jan 03 '17 at 10:47
  • @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. – The Substitute Jan 03 '17 at 11:25
  • If $a_n>0$ and $\Sigma a_n=1$ then $na_n$ must to go zero just FYI – jimjim Jan 04 '17 at 12:41
  • @Arjang I think that only holds when the sequence $(a_n)$ is non-increasing. For example, if $a_n=1/2^n$ for $n$ not a power of $2$ and $a_{2^n}=1/2^n$, then $\sum a_n$ converges, but for $n=2^k$ we have $na_n=2^k*(1/2^k)=1$, so that the sequence $na_n$ doesn't go to zero. – The Substitute Jan 04 '17 at 13:19
  • @TheSubstitute : if $n=2^k$ then $\Sigma a_n \neq 1$, for $\Sigma a_n = 1$ and $a_n>0$ then from some point onwards $a_n$ must be decreasing. – jimjim Jan 04 '17 at 21:03
  • @Arjang I can normalize to make it $1$. Here's a counter to your last statement. Let $n\ge 1$. Take $a_n=(3/2)/2^n$ when $n$ is even and $a_n= (4/3)/3^n$ when $n$ is odd. Then $\sum_{n\ge 1} a_n= (3/2)(1/2^2+1/2^4+\cdots)+(4/3)(1/3+1/3^3+\cdots)=1$. However, $a_{2n}>a_{2n-1}$ for $n>1$ so that $a_n$ isn't eventually decreasing. – The Substitute Jan 05 '17 at 01:29
  • @TheSubstitute : that is example of it not being monotonically decreasing, but it is decreasing and $na_n$ must go to zero. There is a difference between monotonically decreasing and decreasing. still from some points onwards every term must be less then some $\epsilon$ whatever that $\epsilon$ maybe, since we are dealing with positive terms only. The main point is this if $\sum a_n=1$ and $a_n>0$ then $na_n \to 0$ and there exists an $m$ s.t. if $n>m$ the $a_m>a_n$ – jimjim Jan 05 '17 at 02:30
  • @Arjang Ok, well the earlier example I gave ($a_n=1/2^n$ if $n$ not a power of $2$ and $a_{2^k}=1/2^k$), after normalizing, gives a series of positive terms whose sum converges to $1$ yet does not satisfy $na_n\to 1$ because $2^k*a_{2^k}=2^k/2^k=1$ for all $k$. In order to make your claim, you need to be given that $a_n$ is eventually decreasing monotonically - which need not happen by my other counterexample that satisfies $a_{2n}>a_{2n-1}$ for $n\ge 2$. For another reference, see (http://planetmath.org/sites/default/files/texpdf/41941.pdf) – The Substitute Jan 05 '17 at 04:10
  • @TheSubstitute : I can not see how it is possible to have sum of series of positive terms converging where $na_n \not \to 0$, thanks for the refs I check them out – jimjim Jan 05 '17 at 05:31
  • Let us continue this discussion in chat. – The Substitute Jan 05 '17 at 05:32

1 Answers1

1

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.
ajotatxe
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