$\displaystyle \sum_{n=1}^\infty ({ a^{1/n} - \frac{b^{1/n}+c^{1/n}}{2}}),\ if (a>0,b>0,c>0)$
Edit 1
Sorry, replaced i with n. Obviously, term test is inconclusive due to limit is equal to zero. Actually, i tried ratio test several times but didn't succeed and got 1 as a result every time. Also tried to represent a,b,c as exponents and expand them in Taylor series but it didn't work too.
On
Thank's a lot @Mindlack for help. For all who didn't understand above-stated hint here is the solution:
$\sum_{n=1}^\infty ({ a^{1/n} - \frac{b^{1/n}+c^{1/n}}{2}})=\sum_{n=1}^\infty(e^{\frac{ln(a)}{n}}-\frac{e^{\frac{ln(b)}{n}}+e^{\frac{ln(c)}{n}}}{2})$
Expanding exponential function as Mclaurin series and using the logarithm properties:
$\sum_{n=1}^\infty\frac{(2+2\left(\frac{ln(a)}{n}\right)-(1+\left(\frac{ln(b)}{n}\right)+1+\left(\frac{ln(c)}{n}\right))+O(n^{-2}))}{2}=\sum_{n=1}^\infty\frac{ln(\frac{a^2}{bc})}{2n}+O(n^{-2})$
The second term of sum converges everywhere while the first term is harmonic series which diverges everywhere except $a=\sqrt{bc}$.
Answer: diverges everywhere except $a=\sqrt{bc}$.
Hint: $x^{1/n}=1+\frac{\ln(x)}{n} + O(n^{-2})$.