$\liminf X_n^{1/n}<1$ would it imply $\sum_{n=1}^{\infty}X_n$ covergent, where $(X_n)$ is positive sequence.
If not please help me in finding two examples $\liminf X_n^{1/n}<1$ and series convergent and in second example it divergent.
Note: If $\limsup X_n^{1/n}<1$ then the series is convergent and if $\liminf X_n^{1/n}>1$ series divergent.
Further Question: Can we find above two examples keeping $\limsup X_n^{1/n}=1$
If you take $a_{2k}=2^{-k}$ and $a_{2k+1}=2^k$ then $$\liminf_{n\to\infty} a_n^{1/n}=\lim_{k\to\infty} 2^{-k/2k}=\frac{1}{\sqrt{2}}<1$$ but $$\sum_{n=1}^{\infty}a_n=\infty$$
The $\limsup$ is needed in the root test. In this case $$\limsup_{n\to\infty} a_n^{1/n}=\lim_{k\to\infty}2^{\frac{k}{2k+1}}=\sqrt{2}>1$$