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Question: Let \(\displaystyle \{ a_n \}_{n=0}^{\infty}\) be a sequence of positive real numbers. Prove that if \(\sum_{n=0}^{\infty} a_n\) converges, then so does \(\sum_{n=0}^{\infty} {a_n}^r,\) where \(r\) is a positive integer.

 

What I have tried:

I know that if  \(\sum_{n=0}^{\infty} a_n\) converges, it means that \(\displaystyle \{ a_n \}\) goes to 0, which is important because it tell us that when \(a_n\) is less than 1, \({a_n}^{r}\) is also less than 1. 

Also, the fact that \(a_n\) approaches 0 as \(n\) approaches infinity allows us to conclude that there exists a positive integer \(N\) for which \(a_n < 1\) for all \(n \ge N\), meaning that \({a_n}^r\) is also less than 1. 

 

 

Not sure what to do from here, though. :(

 Jun 27, 2022
 #1
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This is so EZ.  Just use the nth term test.

 Jun 27, 2022
 #2
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I am also having the same Issue. Did you get any solution?

 

 

 

 

MyBalanceNow

 Jun 28, 2022

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