I got a question which is

Suppose that $a_n \ge 0 $ $\forall n \in \mathbb{N}$ and that $\sum^{\infty}_{n=1}a_n$ converges. Prove that $\sum^{\infty}_{n=1}(a_n)^2$ also converges.

And what I did is

Given that $\sum^{\infty}_{n=1}a_n$ converges than $lim_{n\to\infty}(a_n)=0$, the there exists $N \in \mathbb{N}$ such that $$0 \le a_n < 1, \quad\forall n \ge N$$

so this means$$0<a_n^2\leq a_n<1,\quad\forall n\geq N$$

but then I am not sure if I can apply the Comparison Test straight away, because I learnt the Comparison Test as

Suppose $b_n \ge a_n \ge 0$ $\forall n \in \mathbb{N}$, then if $\sum^{\infty}_{n=1}b_n$ converges then so does $\sum^{\infty}_{n=1}a_n$.

Whereas in this case I have only proved the inequality for $n \ge N$, can I still apply the Comparison Test and say that since $$0<a_n^2\leq a_n<1,\quad\forall n\geq N$$ then by the Comparison Test, $\sum^{\infty}_{n=1}(a_n)^2$ also converges. Or if there are more steps that I need to include before I make the conclusion.

Thanks to anybody who helps.

  • $\begingroup$ Discarding a finite number of terms does not change the convergence. $\endgroup$ – M. Strochyk Nov 23 '14 at 20:45
  • $\begingroup$ As the previous comment indicates, what you did is fine, since $\sum_{n=1}^{\infty}a_n$ converges iff $\sum_{n=N}^{\infty}a_n$ converges. $\endgroup$ – user84413 Nov 24 '14 at 1:21

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