3
$\begingroup$

I am self studying real analysis.I have come up with following proof and I know that other proofs exist. But, Can someone just tell me if there is anything wrong with the following proof.Thanks in advance

1.we will show that it satisfies cauchy criterion for series

2.consider arbitrary $\epsilon > 0$

3.since we know that $\sum a_n$ converges absolutely

we know that there exists a N s.t forall $n > m \geq N$ s.t $|\sum_{k=m+1}^{n} |a_k|| = \sum_{k=m+1}^{n} |a_k| < \sqrt{\epsilon}$

4.we know show that this N indeed suffices

5.consider arbitrary $n > m \geq N$

$|\sum_{k=m+1}^{n} a_k^2| = \sum_{k=m+1}^{n} a_k^2 \leq (\sum_{k=m+1}^{n} |a_k|)^2 < \sqrt{\epsilon}^2 = \epsilon$

6.so conclude that $\sum a_n^2$ converges

$\endgroup$

marked as duplicate by rtybase, Lord Shark the Unknown, Xander Henderson, Tianlalu, Leucippus May 24 at 5:26

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

10
$\begingroup$

Your proof looks OK. However, there is an (I think) easier and (I strongly think) more intuitive way.

  1. There exists some $N$ such that $|a_n|<1$ for all $n>N$.
  2. The sum $$\sum_{n=1}^\infty a_n^2$$ converges if and only if $$\sum_{n=N}^\infty a_n^2$$ converges.
  3. For the second sum, we have $a_n^2 =|a_n|^2<|a_n|$ for all $n>N$.
  4. Therefore, by comparison test, the sum of $a_n^2$ converges.
$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.