# An inequality between series

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...

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*}
• Agreed and +1 for switching the main variable of summation from $k$ to $j$. – punctured dusk Jan 31 '13 at 16:51
• @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/… – user940 Jan 31 '13 at 16:56
• Seems like the second-to-last step is wrong. $$\sum_{k=j}^\infty \left(1/k^2-1/(k+1)^2\right) = 1/j^2$$ – Thomas Andrews Jan 31 '13 at 17:05