Examine convergence of $\sum_{n=1}^{\infty}(\sqrt[n]{a} - \frac{\sqrt[n]{b}+\sqrt[n]{c}}{2})$

Question

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?

Answer 1

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 $$ a^2=bc. $$

Answer 2

You have, from Taylor expansion of $e^u$ around $0$ to second order, $$\begin{align} \sqrt[n]{a} - \frac{\sqrt[n]{b}+\sqrt[n]{c}}{2} &= e^{\frac{1}{n}\ln a} - \frac{1}{2}\left(e^{\frac{1}{n}\ln c}+e^{\frac{1}{n}\ln b}\right)\\ &= 1+\frac{1}{n}\ln a + \frac{1}{2n^2}\ln ^2 a \\&\qquad- \frac{1}{2}\left(2+\frac{1}{n}\ln b + \frac{1}{2n^2}\ln ^2 b+\frac{1}{n}\ln c + \frac{1}{2n^2}\ln ^2 c\right) + o\left(\frac{1}{n^2}\right) \\ &= \frac{1}{2n}(2\ln a - (\ln b + \ln c)) + \frac{1}{4n^2}(2\ln ^2 a - ( \ln ^2 b+\ln ^2 c)) + o\left(\frac{1}{n^2}\right). \end{align}$$

If $(2\ln a - (\ln b + \ln c)) \neq 0$, can you conclude (by theorems of comparisons)? And if $(2\ln a - (\ln b + \ln c)) = 0$?