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How can one prove for any sequence of positive numbers $a_n, n\ge1,$ we have $$\sum_{n=1}^\infty \frac{n}{a_1+a_2+a_3+\cdots+a_n}\le 2\sum_{n=1}^\infty \frac{1}{a_n}$$


Added later:

Apparently, this is a version of Hardy's inequality. The above is the case $p=-1$. (See the wiki for what $p$ is).

The case $p=2$ appears here: Proving $A: l_2 \to l_2$ is a bounded operator

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    $\begingroup$ Source? Motivation? Failed approaches? $\endgroup$
    – Did
    Commented Jan 12, 2012 at 16:15
  • $\begingroup$ Am I the only one that doesn't get the $1^n/a_1$ in the question? Or pretty much the entire second line for that matter? $\endgroup$
    – anon
    Commented Feb 13, 2012 at 5:05
  • $\begingroup$ @anon: That is an idea OP had. It was an answer, which was moved into the question by Zev. $\endgroup$
    – Aryabhata
    Commented Feb 13, 2012 at 5:51

3 Answers 3

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Here is a proof.

We try using induction, but as usual, a direct approach seems to fail and we have to try and prove a stronger statement.

So we try and pick a positive function $f(n)$ such that

$$ \frac{f(n)}{a_1 + a_2 + \dots + a_n} + \sum_{k=1}^{n} \frac{k}{a_1 + a_2 + \dots + a_k} \le \sum_{j=1}^{n} \frac{2}{a_j}$$

Let $S = a_1 + a_2 + \dots + a_n$ and let $x = a_{n+1}$.

In order to prove that $n$ implies $n+1$ it would be sufficient to prove

$$\frac{2}{x} + \frac{f(n)}{S} \ge \frac{f(n+1) + n+1}{S+x}$$

This can be rearranged to

$$2S^2 + f(n) x^2 + (f(n) + 2) Sx \ge (f(n+1) + n+1) Sx$$

Since $$ 2S^2 + f(n) x^2 \ge 2 \sqrt{2 f(n)} Sx$$

it is sufficient to prove that $f(n)$ satisfies

$$ f(n) + 2 + 2\sqrt{2 f(n)} \ge f(n+1) + n + 1$$

Choosing $f(n) = \dfrac{n^2}{2}$ does the trick.

We can easily verify the base case for this choice of $f(n)$.

Thus we have:

$$ \frac{n^2}{2(a_1 + a_2 + \dots + a_n)} + \sum_{k=1}^{n} \frac{k}{a_1 + a_2 + \dots + a_k} \le \sum_{j=1}^{n} \frac{2}{a_j}$$

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  • $\begingroup$ Nice solution! With $a_n=2^n$ we can see that the constant $2$ cannot be improved. $\endgroup$ Commented Feb 12, 2012 at 20:29
  • $\begingroup$ @DavideGiraudo: Doesn't $a_n = 2^n$ give us the constant to be ~$1.37$? Wolfram Alpha link: wolframalpha.com/input/?i=sum+n%2F%282%282^n+-+1%29%29 $\endgroup$
    – Aryabhata
    Commented Feb 12, 2012 at 20:42
  • $\begingroup$ @DavideGiraudo: btw, I forgot to say: Thank you! $\endgroup$
    – Aryabhata
    Commented Feb 12, 2012 at 21:01
  • $\begingroup$ @DavideGiraudo: Setting $a_1 = 1$, $a_2 = 1$ and $a_n = 2^{n-2}$ seems to work though. $\endgroup$
    – Aryabhata
    Commented Feb 12, 2012 at 22:01
  • $\begingroup$ Indeed, I make mistakes in the computations. $\endgroup$ Commented Feb 12, 2012 at 22:10
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There is also the following N.Sedrakyan's proof.

By C-S $$2\sum_{k=1}^n\frac{1}{a_k}>2\sum_{k=1}^n\frac{k^2}{a_k}\left(\frac{1}{k^2}-\frac{1}{(n+1)^2}\right)=2\sum_{k=1}^n\frac{k^2}{a_k}\sum_{m=k}^n\left(\frac{1}{m^2}-\frac{1}{(m+1)^2}\right)=$$ $$=2\sum_{k=1}^n\left(\frac{1}{k^2}-\frac{1}{(k+1)^2}\right)\sum_{m=1}^k\frac{m^2}{a_m}=2\sum_{k=1}^n\left(\frac{2k+1}{k^2(k+1)^2\sum\limits_{m=1}^ka_m}\sum\limits_{m=1}^ka_m\sum_{m=1}^k\frac{m^2}{a_m}\right)\geq$$ $$\geq2\sum_{k=1}^n\left(\frac{2k+1}{k^2(k+1)^2\sum\limits_{m=1}^ka_m}\left(\sum\limits_{m=1}^km\right)^2\right)=2\sum_{k=1}^n\frac{(2k+1)k^2(k+1)^2}{4k^2(k+1)^2\sum\limits_{m=1}^ka_m}>\sum_{k=1}^n\frac{k}{\sum\limits_{m=1}^ka_m}.$$

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    $\begingroup$ There's a typo with a factor 2 in the 2nd term. $\endgroup$ Commented Oct 4, 2019 at 0:49
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Forgive me as it is my first time on math exchange and I dont know how formatting is done for expressions.

first thing I feel is factor of 2 on right side is not necessary. It can be any number greater than or equal to 1. Equality can occur only when all a_i are identical and that factor on right side is 1.

lets call expression on right side as RHS and on left side as LHS.

Define f(n) = RHS_n - LHS_n
then f(n+1) = RHS_{n+1} - LHS_{n+1}

it is easy to see that

f(n+1) - f(n)>0 for all n

(final expression will be quadratic in a_{n+1} and X, where X = summation of a_i [i =1 to n]).

given inequality is obvious when n =1, so that f(1) >0. hence f(n) > 0 for all n.

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  • $\begingroup$ Are you saying that $\frac{n}{a_1+a_2+a_3+\cdots+a_n}\le\frac{1}{a_n}$ for every $n$? Or maybe that $\frac{n}{a_1+a_2+a_3+\cdots+a_n}\le\frac{2}{a_n}$ for every $n$? This is not clear to me. $\endgroup$
    – Did
    Commented Jan 13, 2012 at 15:46
  • $\begingroup$ Let $A_n=a_1+a_2+a_3+\cdots+a_n$. The unique sequence $(a_n)$ such that $A_n=2^n$ for every $n$ shows that the factor $2$ on the RHS cannot be replaced by anything less than $\frac43$, and in particular not by $1$. $\endgroup$
    – Did
    Commented Jan 13, 2012 at 15:49
  • $\begingroup$ @DidierPiau, No. I am not considering inequality mentioned in your first comment. There you are comparing each term of sequence where as I am considering their sum. $\endgroup$
    – chatur
    Commented Jan 16, 2012 at 5:17
  • $\begingroup$ Then what is the meaning of the assertion that $f(n+1)-f(n)\gt0$ for all $n$? And what about your statement that the factor $2$ on the RHS can be replaced by $1$? $\endgroup$
    – Did
    Commented Jan 16, 2012 at 6:24
  • $\begingroup$ f(n+1) stands for difference between RHS and LHS. where RHS, LHS are themselves sum of terms from index 1 to n+1. If all the terms a_i are equal then we can see that respective terms on right and left side are equal, so is their sum. Generally in inequality, two sides are equal only when all the a_i are equal.(this is not a rule though, it is just my feeling). Probably I have not presented my arguments well enough or there is some flaw in arguments. I will go through it again, though thanks for reading and feedback of yours. $\endgroup$
    – chatur
    Commented Jan 16, 2012 at 6:42

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