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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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Do you mean $x_n$ or $a_n$? And is $a_{n-1}<a_n$? –  draks ... Jan 12 '12 at 15:52
    
i mean $a_n$ , sorry! and nothing is known about the monotony –  FrConnection Jan 12 '12 at 16:00
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Source? Motivation? Failed approaches? –  Did Jan 12 '12 at 16:15
    
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? –  anon Feb 13 '12 at 5:05
    
@anon: That is an idea OP had. It was an answer, which was moved into the question by Zev. –  Aryabhata Feb 13 '12 at 5:51

2 Answers 2

up vote 3 down vote accepted

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:

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

In the infinite summation case, the constant $2$ is the best we can do, as we can see by setting $a_1 = 1$ and $a_n = 2^{n-2}$ for $n \gt 1$.

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Nice solution! With $a_n=2^n$ we can see that the constant $2$ cannot be improved. –  Davide Giraudo Feb 12 '12 at 20:29
    
@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 –  Aryabhata Feb 12 '12 at 20:42
    
@DavideGiraudo: btw, I forgot to say: Thank you! –  Aryabhata Feb 12 '12 at 21:01
    
@DavideGiraudo: Setting $a_1 = 1$, $a_2 = 1$ and $a_n = 2^{n-2}$ seems to work though. –  Aryabhata Feb 12 '12 at 22:01
    
Indeed, I make mistakes in the computations. –  Davide Giraudo Feb 12 '12 at 22:10

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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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. –  Did Jan 13 '12 at 15:46
    
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 $1$. –  Did Jan 13 '12 at 15:49
    
@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. –  chatur Jan 16 '12 at 5:17
    
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 $1$? –  Did Jan 16 '12 at 6:24
    
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. –  chatur Jan 16 '12 at 6:42

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