Question

The following advanced exercise use a generalized ratio test to determine convergence of some series that arise in particular applications, including

Bivariate numerical data
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asked 2021-06-16

The following advanced exercise use a generalized ratio test to determine convergence of some series that arise in particular applications, including the ratio and root test, are not powerful enough to determine their convergence. The test states that if $ \(\displaystyle\lim{\left\lbrace{n}\rightarrow\infty\right\rbrace}{\frac{{{a}{\left\lbrace{2}{n}\right\rbrace}}}{{{a}_{{{n}}}}}}{<}\frac{{1}}{{2}}\) then \(\displaystyle\sum{a}_{{{n}}}\) converges, while if \(\displaystyle\lim{\left\lbrace{n}\rightarrow\infty\right\rbrace}{\frac{{{a}{\left\lbrace{2}{n}+{1}\right\rbrace}}}{{{a}_{{{n}}}}}}{>}\frac{{1}}{{2}}\) then \(\displaystyle\sum{a}_{{{n}}}\) $ diverges. Let \(\displaystyle{a}_{{{n}}}={\frac{{{n}^{{{\ln{{n}}}}}}}{{{\left({\ln{{n}}}\right)}^{{{n}}}}}}\). Show that \(\frac{a_{2n}}{a_{n}} \rightarrow 0\) as \(\displaystyle{n}\rightarrow\infty\)

Expert Answers (1)

2021-06-17
Proved, \(\displaystyle\lim_{{{n}\rightarrow\infty}}{\frac{{{a}_{{{2}{n}}}}}{{{a}_{{{n}}}}}}={0}\)
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