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?
How can one determine if a convergent series raised to a power is also convergent?
4 Answers
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.
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0I understood the latter point; perhaps I should have made the context more clear above -- I'm studying metric spaces and $l_p$ norms, and the instructor sometimes seemed to use the above interchangeably. – 2011-02-02
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 $s = 1$.
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0@bosmacs: Thanks, but I honestly think that @Ross Millikan's answer is by far the best of the three. – 2011-02-02
Denote $S=\sum_{i=1}^{\infty}|x_i|^p$. When does $S^{\frac{1}{p}}$ exist?
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0Unless all $x_i=0$, S>0. And $S^{\frac{1}{p}}$ exists for $p\ne 0$. – 2011-02-03
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 then the inequality is reversed.