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Proving Convergence

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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