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Let $(a_n)_n$ be a strictly positive sequence . How to prove that the series $$ \sum\limits_{n = 1}^\infty {\frac{{a_n }}{{(a_1 + \cdots + a_n )^2 }}} $$ converges ? Any ideas ?

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  • $\begingroup$ $\sum_{n=1}^\infty \frac{a_n}{a_n^2}$ is an upper bound. $\endgroup$ – davcha Oct 28 '14 at 23:07
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Keyword: Telescoping series.

For every $n\geqslant1$, let $A_n=a_1+\cdots+a_n$, then, for every $n\geqslant2$, $A_n\geqslant A_{n-1}$ hence $$\frac{a_n}{A_n^2}\leqslant\frac{a_n}{A_nA_{n-1}}=\frac1{A_{n-1}}-\frac1{A_n}.$$ Summing these yields, for every $N$, $$\sum_{n=1}^N\frac{a_n}{A_n^2}\leqslant\frac1{a_1}+\sum_{n=2}^N\frac1{A_{n-1}}-\frac1{A_n}=\frac2{a_1}-\frac1{A_N}\leqslant\frac2{a_1}.$$ Thus, the series in the question converges and $$\sum_{n=1}^\infty\frac{a_n}{A_n^2}\leqslant\frac2{a_1}.$$

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    $\begingroup$ That's kickass :) $\endgroup$ – Greg Martin Oct 29 '14 at 0:27
  • $\begingroup$ Nice one:) . I would have used comparison with integral, it can even give you the inf of {a; $\sum \frac{a_n}{{S_n}^a}$ converges} $\endgroup$ – mvggz Oct 30 '14 at 10:01
  • $\begingroup$ How can we get the value of the best possible constant $a_1$ ? $\endgroup$ – Victor Oct 30 '14 at 19:07
  • $\begingroup$ @Victor Sorry? Note that $a_1$ is given a priori, as every other $a_n$. $\endgroup$ – Did Oct 30 '14 at 21:33
  • $\begingroup$ THAT'S EXACTLY WHAT I THOUGHT at first !!!I presented my calculus professor this interesting problem as well as your elegant solution.He then replied with the following question: 'If $a_1=1$ is $1$ the best possible constant to bound the series?If so why?Hint, think of Riemann sums.' The Riemann sum bit left me in the darkness completely.. $\endgroup$ – Victor Oct 30 '14 at 21:39
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First note that we can assume all the $a_n$ are at most $1$. If not, replace the term $a_n$ with $\lfloor a_n\rfloor$ copies of $1$ and a leftover $a_n-\lfloor a_n\rfloor$; this only increases the series.

Define $k(m)$ to be the least index $k$ such that $a_1+\cdots+a_k \ge m$ (with $k(0)=1$). Then \begin{align*} \sum_{n=1}^\infty \frac{a_n}{(a_1+\cdots+a_n)^2} &= \sum_{m=0}^\infty \sum_{n=k(m)}^{k(m+1)-1} \frac{a_n}{(a_1+\cdots+a_n)^2} \\ &\le \sum_{m=0}^\infty \sum_{n=k(m)}^{k(m+1)-1} \frac{a_n}{m^2} \\ &= \sum_{m=0}^\infty \frac1{m^2} \sum_{n=k(m)}^{k(m+1)-1} a_n \\ &\le \sum_{m=0}^\infty \frac1{m^2} \cdot2 = \frac{\pi^2}3. \end{align*} The fact that $\sum_{n=k(m)}^{k(m+1)-1} a_n \le 2$ comes from the definition of the $k(m)$ and the fact that the $a_n\le1$.

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