TITLE: Examine convergence of $\sum_{n=1}^{\infty}(\sqrt[n]{a} - \frac{\sqrt[n]{b}+\sqrt[n]{c}}{2})$
QUESTION [7 upvotes]: How to examine convergence of $\sum_{n=1}^{\infty}(\sqrt[n]{a} - \frac{\sqrt[n]{b}+\sqrt[n]{c}}{2})$ for $a, b, c> 0$ using Taylor's theorem?

REPLY [3 votes]: You have, using Taylor's polynomial,
$$
a^{1/n}=e^{\frac1n\,\log a}=1+\frac1n\log a+\frac{e^{c_n}}{2n^2}\,\log^2 a,
$$
where $0<c_n<\frac1n\,\log a$. So 
\begin{align}
\sum_{n=1}^{\infty}(\sqrt[n]{a} - \frac{\sqrt[n]{b}+\sqrt[n]{c}}{2})
&=\sum_{n=1}^\infty\frac1n\,\left(\log a-\frac12\log b-\frac12\,\log c\right)+\frac{1}{2n^2}\left(e^{c_n}\log^2 a-\frac{e^{d_n}}2\log^2b-\frac{e^{f_n}}2\,\log^2c\right)\\ \ \\
&=\sum_{n=1}^\infty\frac1n\,\left(\log \frac a{(bc)^{1/2}}\right)+\frac{1}{2n^2}\left(e^{c_n}\log^2 a-\frac{e^{d_n}}2\log^2b-\frac{e^{f_n}}2\,\log^2c\right).\\ \ \\
\end{align}
The series of the second terms will always converge because of the $n^2$ and the fact that the exponentials are bounded by fixed numbers. So the convergence is decided by $$\sum_{n=1}^\infty\frac1n\,\left(\log \frac a{(bc)^{1/2}}\right).$$
If the log is any nonzero number, you get a divergent series. Thus convergence happens if and only if $a/(bc)^{1/2}=1$, that is
$$
a^2=bc.
$$