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?

  • $\begingroup$ you just asked the same question here $\endgroup$ – reuns Jan 24 '17 at 13:21
  • $\begingroup$ @user1952009: No, it's a different one. $\endgroup$ – Alex M. Jan 24 '17 at 13:21
  • $\begingroup$ 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 $\endgroup$ – Euler_Salter Jan 24 '17 at 13:22
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    $\begingroup$ @Euler_Salter: I really don't understand how you get that $\sum_{n=1}^ka_n^s < \left(\sum_{n=1}^ka_n\right)^2$. $\endgroup$ – Alex M. Jan 24 '17 at 13:22
  • $\begingroup$ 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 $\endgroup$ – reuns Jan 24 '17 at 13:23

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


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.

  • $\begingroup$ after you say "hence for some $m$ we have.." do we have $a_n < 1$ because we fixed "$\epsilon$" to be $1$? $\endgroup$ – Euler_Salter Jan 24 '17 at 13:30
  • $\begingroup$ oh okay and then use comparison test of I can just used Completeness axiom. Yeah this is kind of the proof of the book $\endgroup$ – Euler_Salter Jan 24 '17 at 13:31
  • $\begingroup$ I've changed the typo now $\endgroup$ – Euler_Salter Jan 24 '17 at 13:39

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