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Assume that $\displaystyle\sum_{n=1}^\infty a_n$ is a convergent series with $a_n>0, n=1,2\dots$

How can we prove that $\displaystyle\sum_{n=1}^\infty (a_n)^{\frac{\ln n}{1+\ln n}}$ is also a convergent series?


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Define $$I=\{n:a^{\ln{n}/(1+\ln{n})}_{n}\le e^2a_{n}\},J=N\I$$ if $n\in J$,then $$a^{\ln{n}}_{n}>e^{2+2\ln{n}} a^{1+\ln{n}}_{n}=(en)^2\cdot a^{1+\ln{n}}_{n},\Longrightarrow a_{n}<(en)^{-2}$$ therefore $$\sum_{n=1}^{\infty}a^{\dfrac{\ln{n}}{1+\ln{n}}}_{n}\le \sum_{n\in I}e^2a_{n}+\sum_{n\in J}(en)^{-2}<\infty$$

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