# if $\sum_{n=1}^{\infty} a_n$ converges absolutely then prove that $\sum_{n=1}^{\infty} a_n^2$ converges [duplicate]

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

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

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.