级数与极限一题

设正项级数$\sum\limits_{n=1}^{\infty} a_n$收敛,证明:极限$\cfrac{n^2}{\frac{1}{a_1}+\frac{1}{a_2}+\cdots+\frac{1}{a_n}}$存在.
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妖心儿

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$n\ge m,\\
\cfrac {n^2}{\frac 1{a_1}+\frac 1{a_2}+\cdots+\frac 1{a_n}}=\cfrac {n^2}{(n-m+1)^2}\cfrac {(n-m+1)^2}{(\frac 1{a_1}+\cdots+\frac 1{a_m})+\frac 1{a_{m+1}}+\cdots+\frac 1{a_n}}\\
\le\cfrac {n^2}{(n-m+1)^2}(\cfrac 1{\frac 1{a_1}+\cdots+\frac 1{a_m}}+a_{m+1}+\cdots+a_n)\\
\le\cfrac {n^2}{(n-m+1)^2}(a_m+a_{m+1}+\cdots+a_n).\\
n\to\infty,\\
\limsup\limits_{n\to\infty}\cfrac {n^2}{\frac 1{a_1}+\frac 1{a_2}+\cdots+\frac 1{a_n}}\le\sum\limits_{k=m}^{\infty} a_k.\\
m\to\infty\Rightarrow\lim\limits_{n\to\infty}\cfrac {n^2}{\frac 1{a_1}+\frac 1{a_2}+\cdots+\frac 1{a_n}}=0.
$

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