Problem
$$ 已知正项级数\sum_{n=1}^{\infty}{a_n}收敛，证明：\sum_{n=1}^{\infty}{a_{n}^{\frac{n}{n+1}}}收敛. $$
Tips Answer
Tips:
$$
分别考虑a_{n}^{\frac{n}{n+1}}\le 2a_n与a_{n}^{\frac{n}{n+1}}>2a_n两种情况.
$$
Answer:
$$ \begin{aligned} \sum_{n=1}^{\infty}{a_{n}^{\frac{n}{n+1}}}&=\sum_{a_{n}^{\frac{n}{n+1}}\le 2a_n}{a_{n}^{\frac{n}{n+1}}}+\sum_{a_{n}^{\frac{n}{n+1}}>2a_n}{a_{n}^{\frac{n}{n+1}}}\\ &\le 2\sum_{n=1}^{\infty}{a_n}+\sum_{a_{n}^{\frac{1}{n+1}}<\frac{1}{2}}{a_{n}^{\frac{n}{n+1}}}=2\sum_{n=1}^{\infty}{a_n}+\sum_{a_{n}^{\frac{n}{n+1}}<\frac{1}{2^n}}{a_{n}^{\frac{n}{n+1}}}\\ &\le 2\sum_{n=1}^{\infty}{a_n}+\sum_{n=1}^{\infty}{\frac{1}{2^n}}<\infty\\ \end{aligned} $$
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