2
$\begingroup$

Given a sequence $(a_n)_{n=1}^\infty$ of positive reals. How do I prove that

$$\sum_{n=1}^\infty \frac{n}{a_1 + \ldots + a_n}\leqslant 2 \sum_{n=1}^\infty \frac{1}{a_n}$$

Of course if the right hand side converges, then $a_n$ is eventually increasing to $\infty$ but the difficulty for me arises from the fact that the behaviour of some first finite number of terms can be arbitrary...

$\endgroup$
3
$\begingroup$

This is based on Grahame Bennett's solution to American Mathematical Monthly problem 11145 published in April 2005. The solution appeared in the October 2006 issue.

The Cauchy-Schwarz inequality gives $(\sum_1^k j)^2\leq \sum^k_1j^2/a_j\, \sum^k_1 a_j$, or equivalently, $${k\over\sum_{j=1}^k a_j}\leq{4\over k(k+1)^2}\sum_{j=1}^k {j^2\over a_j}.$$ Summing over $k$ yields \begin{eqnarray*} \sum_{k=1}^\infty \frac{k}{a_1 + \cdots + a_k}&\leq&2 \sum_{j=1}^\infty{j^2\over a_j}\sum_{k=j}^\infty {2\over k(k+1)^2}\leq 2 \sum_{j=1}^\infty {j^2\over a_j}\sum_{k=j}^\infty{2k+1\over k^2(k+1)^2} \\[5pt] & = & 2 \sum_{j=1}^\infty{j^2\over a_j}\sum_{k=j}^\infty\left({1\over k^2}-{1\over(k+1)^2}\right) = 2 \sum_{j=1}^\infty {1\over a_j}. \end{eqnarray*}

$\endgroup$
  • $\begingroup$ Wow, amazing. Thank you very much. $\endgroup$ – user60253 Jan 31 '13 at 16:50
  • $\begingroup$ Agreed and +1 for switching the main variable of summation from $k$ to $j$. $\endgroup$ – punctured dusk Jan 31 '13 at 16:51
  • $\begingroup$ @user60253 Thanks. All credit goes to Grahame Bennett, however! Your inequality is an example of Hardy's inequality, with $p=-1$. Compare with: math.stackexchange.com/questions/44456/… $\endgroup$ – user940 Jan 31 '13 at 16:56
  • $\begingroup$ Seems like the second-to-last step is wrong. $$\sum_{k=j}^\infty \left(1/k^2-1/(k+1)^2\right) = 1/j^2$$ $\endgroup$ – Thomas Andrews Jan 31 '13 at 17:05
  • $\begingroup$ Yes, I introduced an error. I will correct it, thanks! $\endgroup$ – user940 Jan 31 '13 at 17:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.