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Find the smallest possible value of $\lambda$ such that for all sequences $(a_n)_{ n \in \mathbb{N}}$ of positive reals such that $\sum_{n=1}^{\infty} \frac{1}{a_n}$ converges, the following inequality holds:

$\sum_{n=1}^{\infty} \frac{n}{\sum_{k=1}^{n}a_k} \le \lambda \sum_{n=1}^{\infty} \frac{1}{a_n}$

(*)

It is not hard to show using Cauchy-Schwarz and some simple bounding that $\lambda=2$ works.

However showing that $\lambda \ge 2$ is more difficult. What we need to do is for every $\epsilon>0$ exhibit a sequence $(a_n)$ such that $\sum_{n=1}^{\infty} \frac{1}{a_n}$ converges and

$\frac{\sum_{n=1}^{\infty} \frac{n}{\sum_{k=1}^{n}a_k}}{\sum_{n=1}^{\infty} \frac{1}{a_n}} \ge 2-\epsilon$.

(**)

Note that if we do not impose that $\sum_{n=1}^{\infty} \frac{1}{a_n}$ converges, we have equality of partial sums in (*) with the sequence $s_k$ given by $a_n=k*n$ for all $n$, where $k$ is a constant. So the idea is to construct a sequence $t$ which goes to infinity slightly faster than $s$ and which looks "locally" like $s_k$ for some $k$.

I have constructed a rather complicated sequence $t$ based on this idea but I have not yet worked out the details to prove it works.

Is there a simple construction of $a_n$ which satisfies (**)?

math_lover
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1 Answers1

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Let $S_N=\{a_n^{(N)}\}_{n\geq 1}$ be the sequence defined through $$ a_n^{(N)} = \left\{\begin{array}{rcl}n &\text{if}& n\leq N\\ n(n-N)&\text{if}& n>N.\end{array}\right.$$ Dropping the superscript and setting $A_n=a_1+a_2+\ldots+a_n$ we have $A_n=\frac{n^2+n}{2}$ for any $n\leq N$ and $A_n=\frac{(2n+N+1)(n-N+1)(n-N)}{6}+\frac{N^2+N}{2}\geq\frac{n}{3}(n-N)^2$ if $n>N$. Additionally $$ \sum_{n\geq 1}\frac{1}{a_n} = H_N\left(1+\frac{1}{N}\right)=H_N+O(1) $$ $$ \sum_{n\geq 1}\frac{n}{A_n} = 2\left(H_{N+1}-1\right)+O\left(\sum_{n>N}\frac{n}{n(n-N)^2}\right)=2H_N+O(1)$$ so $\lambda=2$ is the optimal constant. Here $H_N$ stands for $\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{N}=\log(N)+O(1)$ as usual and $\sum_{n>N}\frac{1}{n(n-N)}$ equals $\frac{H_N}{N}$ by telescoping.

Jack D'Aurizio
  • 353,855
  • Oops. Didn't see the easy solution... Darn. – math_lover Dec 16 '17 at 19:30
  • We may also notice that if ${a_n}{n\geq 1}$ is a positive sequence such that $\sum{n\geq 1}\frac{1}{a_n}$ is convergent, by Cesàro-Stolz we have

    $$ \lim_{T\to +\infty}\frac{\sum_{n>T}\frac{n}{A_n}}{\sum_{n>T}\frac{1}{a_n}} =\lim_{T\to +\infty}\frac{T a_T}{A_T}=\lim_{T\to +\infty}\frac{a_{T+1}+T(a_{T+1}-a_T)}{a_{T+1}}=1+\beta$$ as soon as $\frac{a_T}{T^\beta}$ is convergent to a non-zero constant. In the shown example, $\beta=2$.

     – Jack D'Aurizio Dec 16 '17 at 19:43