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I want to show that if $\sum_{n = 1}^{\infty} a_n$ converges, then for $s \geq 1$, also $\sum_{n=1}^{\infty}a_n^s$ converges. (with $a_n >0$ $\forall n$)

My book gives a proof based on the contrapositive of the n-th term test, however I developed a proof but I don't know if it's correct or not.

If $\sum_{n = 1}^{\infty} a_n$ converges, then $S_k = \sum_{n = 1}^{k} a_n$ converges to a limit, say $L$.

Now since the terms are all positive we have that the sequence of partial sums is strictly increasing and every term is positive $$0 < \sum_{n=1}^ka_n^s$$ but also (I think, by looking at $(x+y)^2 > x^2+y^2$ ) we have $$0 < \sum_{n=1}^ka_n^s < \left(\sum_{n=1}^ka_n\right)^s=L^s$$ hence the sequence of partial sums is monotone and bounded so by completeness axiom it converges.

Does it work?

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    you just asked the same [question here](http://math.stackexchange.com/questions/2111557/sum-n-1-inftya-n-converges-with-a-n-0-for-all-n-then-sum-n-1) – 2017-01-24
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    @user1952009: No, it's a different one. – 2017-01-24
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    it is really not the same question.. the problem is different. However in preparation for the exam I am trying to develop alternative proofs for each theorem we've done – 2017-01-24
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    @Euler_Salter: I really don't understand how you get that $\sum_{n=1}^ka_n^s < \left(\sum_{n=1}^ka_n\right)^2$. – 2017-01-24
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    It is really the same question : $0 \le \sum_{n=1}^N b_n a_n \le \sum_{n=1}^N a_n$ so $u_N = \sum_{n=1}^N b_n a_n $ is increasing and bounded – 2017-01-24
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    If $\sum a_n$ converges, then $a_n \to 0$ so that $a_n^s\leq a_n$ for all big enough n – 2017-01-24
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    well that's a guess, but I suppose one can prove it by induction, or by using the binomial theorem. I think the sum of the squares is smaller than the square of the sum – 2017-01-24
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    @user1952009: Indeed, viewing it like that they are essentially identical. Still, I believe that it would help the OP to leave this one open, since he is asking us to verify some work done on his own, not for a proof of his claim. – 2017-01-24
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    @user1952009 it's a different theorem, I don't think it is the same at all – 2017-01-24
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    @Euler_Salter One of the issues in your proof is that you have sum of $a_n^s$ (power $s$), and you then upper bound by the square of the sum (power $2$). What if $s=1.0000001$? – 2017-01-24
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    Suppose that $x^s+y^s <(x+y)^2$ for $x,y >0$ and $s \ge 1$. With $s \to 1+0$ we get $x+y <(x+y)^2$, hence $x+y>1$. Therefore, the inequality $x^s+y^s <(x+y)^2$ is not true, in general. – 2017-01-24
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    that is a misprint!! I mean $s$ of course, now I understand – 2017-01-24
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    when I wrote the power two I made a typo, I mean power $s$ – 2017-01-24
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    The one with power two was just an example, but then I just kept on writing $2$ for some reason. – 2017-01-24

2 Answers 2

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The simplest and most straightforward proof is the following: $a_n\gt 0$ and $\sum_{n\geq 0}a_n$ converges mean that $a_n\to 0$. So beyond a certain rank $N$ we have $0\lt a_n\lt 1$. Now $s\geq 1$ means that beyond $N$ we have $0\lt a_n^s\lt a_n$. So the comparison test tells us that $\sum_{n\geq 0}a_n^s$ converges

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The inequality $\sum_{n=1}^ka_n^s < \left(\sum_{n=1}^ka_n\right)^2$ is not true, in general. Find a counter example !

Since $\sum_{n = 1}^{\infty} a_n$ converges, we have $a_n \to 0$, hence, for some $m$ we have

$a_n <1$ for $n>m$. Therefore, since $s \ge 1$:

$a_n^s \le a_n$ for $n>m$.

Your turn.

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    after you say "hence for some $m$ we have.." do we have $a_n < 1$ because we fixed "$\epsilon$" to be $1$? – 2017-01-24
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    oh okay and then use comparison test of I can just used Completeness axiom. Yeah this is kind of the proof of the book – 2017-01-24
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    I've changed the typo now – 2017-01-24