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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.

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If three parameters bother you, maybe try $\sum \sqrt[n]a-\sqrt[n]b$. $a,b$ need to be close for this not to diverge. Do they have to be equal? – Ross Millikan Mar 15 '12 at 13:53
In Ross's simplified version, try $a=1$, $b=1+\beta$. – Harald Hanche-Olsen Mar 15 '12 at 15:36
up vote 1 down vote accepted

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.
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