# The series $\sum\limits_{n=1}^\infty \frac n{\frac1{a_1}+\frac1{a_2}+\dotsb+\frac1{a_n}}$ is convergent

If a series $\sum\limits_{n=1}^\infty a_n$ is convergent, and $a_n\gt0$...

1. Do not refer to Carleman's inequality or Hardy's inequality, show that the series

$$\sum_{n=1}^\infty \frac n{\frac1{a_1}+\frac1{a_2}+\dotsb+\frac1{a_n}}$$

is also convergent.

2. What is the minimum positive real number $k$ such that the following inequality holds for all convergent series $a_n\gt0$?

$$\sum_{n=1}^\infty \frac n{\frac1{a_1}+\frac1{a_2}+\dotsb+\frac1{a_n}}\le k\sum_{n=1}^\infty a_n$$

3. Does it exist a positive real number $l$ such that

$$\sum_{n=1}^\infty a_n \le l\sum_{n=1}^\infty \frac n{\frac1{a_1}+\frac1{a_2}+\dotsb+\frac1{a_n}}$$

holds?

• @Nathaniel while i was reading something about Carleman's inequality and Hardy's inequality, I thought this problem. – ziang chen Dec 11 '13 at 10:33

Edit One step was flawed in the proof of 1). Many thanks to my friend Luc for pointing this out. I have fixed it, thanks to a lemma that I prove at the end.

Using AM-GM inequality, we see that the general term is not greater than $\sqrt[n]{a_1\cdots a_n}$. Hence Carleman's inequality indeed shows that 2) holds with constant $e$. But we need a significantly different approach if we want to get the optimal constant in this case, since it turns out to be $2$. I found this method by investigating the integral version of the problem first. It is indeed easier (the lemma below is exactly what is tedious in the discrete case, while straightforward in the integral case) to see that $$\int_0^\infty \frac{x}{\int_0^x\frac{1}{f(t)}dt}dx\leq 2\int_0^\infty f(t)dt$$ for every positive measurable function $f$, and that $2$ is optimal.

The answer to 3) is no.

Remark When proving 1), we can assume that $a_k$ is nonincreasing withtout loss of generality, since reordering the sequence does not affect the rhs, and can only increase the lhs when making the sequence nonincreasing. Maybe we coould take advantage of that observation to simplify the argument in 1).

1) We will show that for every positive sequence $a_n$, we have

$$\sum_{n=1}^\infty \frac{n}{\sum_{k=1}^n \frac{1}{a_k}}\leq 2 \sum_{k=1}^\infty a_k$$

By Faulhaber's formula and Cauchy-Schwarz $$\frac{n^2(n+1)^2}{4}=\left( \sum_{k=1}^nk\right)^2=\left(\sum_{k=1}^n k\sqrt{a_k}\frac{1}{\sqrt{a_k}}\right)^2\leq \sum_{k=1}^n k^2a_k\sum_{k=1}^n \frac{1}{a_k}$$ whence $$\frac{n}{\sum_{k=1}^n \frac{1}{a_k}}\leq \frac{4}{n(n+1)^2}\sum_{k=1}^n k^2a_k$$ Therefore $$\sum_{n=1}^\infty\frac{n}{\sum_{k=1}^n \frac{1}{a_k}}\leq 4\sum_{n=1}^\infty\frac{1}{n(n+1)^2} \sum_{k=1}^n k^2a_k$$ (by Fubini) $$=4 \sum_{k=1}^\infty k^2a_k\sum_{n=k}^\infty\frac{1}{n(n+1)^2}$$ (by the lemma below) $$\qquad \leq 4 \sum_{k=1}^\infty k^2a_k\frac{1}{2k^2} =2 \sum_{k=1}^\infty a_k$$

2) We will now show that $2$ is optimal. Assume $C>0$ is such that $$\sum_{n=1}^\infty \frac{n}{\sum_{k=1}^n \frac{1}{a_k}}\leq C \sum_{k=1}^\infty a_k$$ for every positive sequence $a_n$. Then for every $\alpha >1$, consider $a_n=\frac{1}{n^\alpha}$. This yields $$C\sum_{n=1}^\infty \frac{1}{n^\alpha}\geq \sum_{n=1}^\infty\frac{n}{\sum_{k=1}^nk^\alpha}\geq \sum_{n=1}^\infty\frac{n}{\int_{x=1}^{n+1}x^\alpha dx} =(\alpha+1) \sum_{n=1}^\infty \frac{n}{(n+1)^{\alpha+1}-1}$$ Now $$\sum_{n=1}^\infty \frac{n}{(n+1)^{\alpha+1}-1}\geq \sum_{n=1}^\infty \frac{n}{(n+1)^{\alpha+1}} =\sum_{n=1}^\infty \frac{1}{(n+1)^{\alpha}}- \frac{1}{(n+1)^{\alpha+1}}$$ $$=\sum_{n=1}^\infty \left(\frac{1}{n^{\alpha}}- \frac{1}{n^{\alpha+1}}\right) \geq \sum_{n=1}^\infty \left(\frac{1}{n^{\alpha}}- \frac{1}{n^{2}}\right) =\sum_{n=1}^\infty \frac{1}{n^{\alpha}} -\frac{\pi^2}{6}$$ Since $\lim_{\alpha\rightarrow 1^+} \sum_{n=1}^\infty \frac{1}{n^{\alpha}}=\sum_{n=1}^\infty\frac{1}{n}=+\infty$ by monotone convergence, $$\frac{C}{\alpha+1}\geq \frac{\sum_{n=1}^\infty \frac{1}{n^{\alpha}} -\frac{\pi^2}{6}}{\sum_{n=1}^\infty \frac{1}{n^{\alpha}}}\quad\Rightarrow\quad \frac{C}{2}\geq 1$$ by letting $\alpha>1$ tend to $1$. So $C\geq 2$ which proves that $2$ is optimal.

3) If such a constant existed, we would get, by AM-GM inequality $$\sum_{n=1}^\infty a_n\leq C\sum_{n=1}^\infty \frac{n}{\sum_{k=1}^n\frac{1}{a_k}}\leq C\sum_{n=1}^\infty\sqrt[n]{a_1\cdots a_n}$$ for every positive sequence. That is, setting $b_n=\sqrt[n]{a_1\cdots a_n}\iff a_n=\frac{b_n^n}{b_{n-1}^{n-1}}$ $$\sum_{n=1}\left(\frac{b_n}{b_{n-1}}\right)^nb_{n-1}\leq C\sum_{n=1}^\infty b_n$$ for every positive sequence $b_n$. Of course this is true with $C=1$ if $b_n$ is nonincreasing. To get our contradiction, we can construct a converging $\sum b_n$ such that the lhs diverges. For example, set $b_n:=\frac{1}{2^n}$ for every $n$ but $b_n=\frac{4}{2^n}$ for, say, every $n=2^k$. That is: we make a bump at every $n=2^k$ which is large enough to make the lhs diverge, but small enough to keep the rhs converging.

Lemma For every $k\geq 1$ we have $$\sum_{n\geq k}^\infty \frac{1}{n(n+1)^2}\leq \frac{1}{2k^2}$$

Proof Let us consider the sequence $$x_k:= \frac{1}{2k^2} - \sum_{n\geq k}^\infty \frac{1}{n(n+1)^2}$$ Note that $\lim_{k\rightarrow +\infty} x_k=0$. Then compute $$x_{k}-x_{k+1}=\frac{1}{2k^2}-\frac{1}{k(k+1)^2}-\frac{1}{2(k+1)^2}=\frac{1}{2k^2(k+1)^2}\geq 0$$ So $x_k$ is nonincreasing and converges to $0$. Therefore $x_k\geq 0$ for every $k\geq 1$. $\Box$

• Thanks! How to prove your first inequality about positive measurable function $f$, $\int_0^\infty \frac{x}{\int_0^x\frac{1}{f(t)}dt}dx\leq 2\int_0^\infty f(t)dt$ ? – ziang chen Dec 19 '13 at 14:02
• @ziangchen You can follow the same steps. First Cauchy-Schwarz on $x=x\sqrt{f(x)}\frac{1}{\sqrt{f(x)}}$. Then Fubini. – Julien Dec 19 '13 at 14:36
• Hardy's inequality has a condition $p>1$ and you have shown it still holds when $p=-1$. Can we find more such $p$? – username Nov 29 '17 at 12:25
• I guess instead of $\sum_{n=1}^\infty \left(\frac{1}{n^{\alpha}}- \frac{1}{n^{\alpha+1}}\right)$ should be $\sum_{n=\color{red} 2}^\infty\dots$. This change also affects the next formulas. – Alex Ravsky Oct 8 '19 at 12:49