Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Prove that if $\sum a_n$ is a convergent series of nonnegative numbers and $p>1$, then $\sum a_n^p$,converges.

My proof is as follows.

By the theorem which says that if a series $\sum a_n$ converges, then $\lim_{n \rightarrow \infty} = 0$, the sequence $(a_n)$ approaches 0 as n approaches infinity. This implies each $a_n$ is in the form of fraction where the denomenator is greater than numerator. Then, $|a_n^p| \leq a_n$, and $\sum a_n^p$ converges by the comparison test.

Is this valid?? I saw back of my text book and gives different proof!

share|improve this question
add comment

1 Answer 1

up vote 1 down vote accepted

You say $a_n$ is "...in the form of fraction where the denominator is greater than numerator." Your idea is fine, but what you want to say is that $a_n<1$ eventually since $a_n\to 0$, so that $$a_n^{1+\varepsilon}=a_n^{\varepsilon}a_n<a_n$$

for in the interval $0<x<1$, we have $x^{\varepsilon }<1$

Note the absolute value bars are unnecessary, since $a_n\geqslant 0$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.