This series converges or diverges,

[mercedesbenz]

This series converges or diverges, If converges, how to prove it.


$\displaystyle{\sum_{n=1}^{\infty}\frac{1}{n^{\frac{4}{3}}}\Bigg(\frac{1}{1}+\frac{1}{2^{\frac{2}{3}}}+\frac{1}{3^{\frac{2}{3}}} +\frac{1}{4^{\frac{2}{3}}}+\cdots+\frac{1}{n^{\frac{2}{3}}}\Bigg)}$

[M@gpie]

เลือก \[ a_n = \frac{1}{n^{4/3}}\left( \frac{1}{1} + \frac{1}{2^{\frac{2}{3}}} + \frac{1}{3^{\frac{2}{3}}} + ... +\frac{1}{n^{\frac{2}{3}}} \right),~~~~~b_n = \frac{1}{n} \]
ลองไล่ดูดีๆจะพบว่า
\[ \lim_{n \rightarrow \infty} \frac{a_n}{b_n} = \lim_{n \rightarrow \infty} \frac{1}{n}\left[ \left( \frac{1}{n} \right)^{-\frac{2}{3}} + \left( \frac{2}{n} \right)^{-\frac{2}{3}} + \left( \frac{3}{n} \right)^{-\frac{2}{3}} + ... +\left( \frac{n}{n} \right)^{-\frac{2}{3}} \right] = \int_0^1 x^{-\frac{2}{3}}dx = 3 \]

เนื่องจาก ${\displaystyle \sum_{n=1}^{\infty} b_n }$ ลู่ออก ดังนั้น ${\displaystyle \sum_{n=1}^{\infty} a_n }$ ลู่ออกด้วย

[nooonuii]

วิธีของน้อง Magpie สวยดีครับ ผมมีอีกวิธีซึ่งใช้อสมการ AM-GM

$\displaystyle{\frac{1}{n^{\frac{4}{3}}}\Bigg(\frac{1}{1}+\frac{1}{2^{\frac{2}{3}}}+\frac{1}{3^{\frac{2}{3}}} +\frac{1}{4^{\frac{2}{3}}}+\cdots+\frac{1}{n^{\frac{2}{3}}}\Bigg)\geq\frac{n}{n^{\frac{4}{3}}(n!)^{\frac{2}{3n}}}\geq\frac{1}{n} }$