Study the convergence of series: $\sum_{n=1}^{+\infty}\left( \sqrt[n]{a}-\frac{\sqrt[n]{b}+\sqrt[n]{c}}{2} \right)$ depending on $a,b,c>0$. I have no idea how to do it, first time I see series with three parameters.
Convergence of interesting series
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sequences-and-series
convergence-divergence
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0In Ross's simplified version, try $a=1$, $b=1+\beta$. – 2012-03-15
1 Answers
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Introduce $t_n=\sqrt[n]{a}-\frac12\sqrt[n]{b}-\frac12\sqrt[n]{c}$. For every fixed positive $x$, when $n\to\infty$, $ \sqrt[n]{x}=\exp\left(\frac1n\log x\right)=1+\frac1n\log x+O\left(\frac1{n^2}\right). $ Hence $t_n=\frac1n\tau+O\left(\frac1{n^2}\right)$ when $n\to\infty$, with $\tau=\log a-\frac12\log b-\frac12\log c=\frac12\log\left(\frac{a^2}{bc}\right)$. Since the series $\sum\limits_n\frac1n$ is divergent and the series $\sum\limits_n\frac1{n^2}$ is absolutely convergent, two cases may occur:
- If $\tau\ne0$, that is, if $a^2\ne bc$, then the series $\sum\limits_nt_n$ diverges, to $+\infty$ if $a^2\gt bc$ and to $-\infty$ if $a^2\lt bc$.
- If $\tau=0$, that is, if $a^2=bc$, then the series $\sum\limits_nt_n$ converges absolutely.