How can one determine if a convergent series raised to a power is also convergent? Forgive me if this is elementary, but my analysis is quite rusty and I'm struggling to get back up to speed. Given a convergent series $\sum_{i=1}^\infty|x_i|^p$, does it follow that $(\sum_{i=1}^\infty|x_i|^p)^\frac{1}{p}$ also converges? If so, how do they relate, i.e. is one always greater than (or equal) to the other? 
 A: The sum is just a number (if it converges).  Call it S.  If S>0 and $p \ne 0$ you can calculate $S^{(1/p)}$
Maybe what you want to ask if if $\sum_{i=1}^\infty|x_i|$ converges does $\sum_{i=1}^\infty|x_i|^p$ converge?  It will if $p \ge 1$ as the $x_i$ are going to zero and raising them to a power greater than 1 will decrease them.
A: You're just taking the $p$th root of the real number $s = \sum_{n=1}^{\infty} |x_{n}|^{p}$. So if $s \gt 1$ then $s \gt s^{\frac{1}{p}}$ and if $s \lt 1$ then $s \lt s^{\frac{1}{p}}$ finally, equality holds if $s = 1$.
A: Denote $S=\sum_{i=1}^{\infty}|x_i|^p$. When does $S^{\frac{1}{p}}$ exist?
A: For your second question, recall the elementary inequality $(a_1+ \cdots +a_n)^r\leq a_1^r+ \cdots + a_n^r$ valid for any natural number $n$, positive numbers $a_1, \ldots, a_n$ and $0<r<1$. Now just apply this to the partial sums: If $p\geq 1$ then $\left( \sum_{k=1}^{m} |a_k|^p \right) ^{\frac{1}{p}} \leq \sum_{k=1}^m |a_k|$ and then take limits. If $0<p<1$ then the inequality is reversed.  
