令{pn}为一个正实数序列,证明若级数∑(pn^-1)收敛,则级数

$\Sigma$$\cfrac{{n}^{2}}{{(p1+p2+……+pn)}^{2}}$pn
也收敛。
已邀请:
$\cfrac 1{S_{n-1}}-\cfrac 1{S_n}=\cfrac{p_n}{S_nS_{n-1}}\ge\cfrac{p_n}{S_n^2}$
$\cfrac{n^2p_n}{S_n^2}\le\cfrac{n^2}{S_{n-1}}-\cfrac {n^2}{S_n}=\cfrac {(n-1)^2}{S_{n-1}}-\cfrac {n^2}{S_n}+\cfrac {2n-1}{S_{n-1}}$
$S_n(\cfrac {1^2}{a_1}+\cfrac{2^2}{a_2}+\cdots+\cfrac{n^2}{a_n})\ge(1+2+\cdots+n)^2=\cfrac{n^2(n+1)^2}4$
$\cfrac n{S_n}\le\cfrac 4{n(n+1)^2}(\cfrac {1^2}{p_1}+\cfrac{2^2}{p_2}+\cdots+\cfrac{n^2}{p_n})$
$\sum\limits_{n=1}^{\infty}\cfrac n{S_n}\le\sum\limits_{n=1}^{\infty}\cfrac 4{n(n+1)^2}(\cfrac {1^2}{p_1}+\cfrac{2^2}{p_2}+\cdots+\cfrac{n^2}{p_n})=\sum\limits_{n=1}^{\infty}\sum\limits_{m=1}^n\cfrac 4{n(n+1)^2}\cfrac{m^2}{p_m}$
$=\sum\limits_{m=1}^{\infty}\sum\limits_{n=m}^{\infty}\cfrac 4{n(n+1)^2}\cfrac{m^2}{p_m}\le\sum\limits_{m=1}^{\infty}\sum\limits_{n=m}^{\infty}2(\cfrac 1{n^2}-\cfrac 1{(n+1)^2})\cfrac{m^2}{p_m}=2\sum\limits_{m=1}^{\infty} \cfrac1{p_n}$

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