# If $\sum_{n = 1}^{\infty} a_n$ converges, with $a_n >0$, $\forall n$, then for $s \geq 1$, also $\sum_{n=1}^{\infty}a_n^s$ converges

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?

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

• after you say "hence for some $m$ we have.." do we have $a_n < 1$ because we fixed "$\epsilon$" to be $1$? – Euler_Salter Jan 24 '17 at 13:30
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