# If $\sum_{1}^{\infty} a_n$ be absolutely convergent then how to show $\sum_{1}^{\infty} a_n$ is also convergent?

If $\sum_{1}^{\infty} a_n$ be absolutely convergent then how to show $\sum_{1}^{\infty} a_n$ is also convergent?

The proof of this question in my textbook:

Let $s_k := \sum_{1}^{k} a_n$ and $t_k := \sum_{1}^{k} |a_n|$.

Then using m>k, $$|s_m - s_k|= |\sum_{k+1}^{m}a_n| \leq \sum_{k+1}^{m}|a_n| = t_m -t_k \to 0.$$

I am really unsure about the above inequality and how that has been used to prove the theorem ? Also I understand that $|t_m -t_k| < \epsilon$, but I am unsure how this implies that $t_m-t_k \to 0$

• The inequality is called triangle inequality. – Santosh Linkha Dec 17 '16 at 13:18
• The only inequality in that proof is one that follows from the triangle inequality: what isn't clear here? – DonAntonio Dec 17 '16 at 13:18
• @DonAntonio I am not sure why that inequality is true ? A proof would be much appreciated – questiontime Dec 17 '16 at 13:24
• @iheartalgebraa I think that someone dealing with analysis should long time ago have studied that. Anyway, if you google "triangle inequality" I'm sure you'll find millions of sites... – DonAntonio Dec 17 '16 at 13:27
• @iheartalgebraa You are not using it "in the case of series". You are using it "in the case of finite sums". – user228113 Dec 17 '16 at 13:35

The proof of this question in my textbook:

let $s_k := \sum_{1}^{k} a_n$ and $t_k := \sum_{1}^{k} |a_n|$. Then using $m>k$, $$\big|s_m - s_k\big|= \biggr|\sum_{k+1}^{m}a_n\biggr| \leq \sum_{k+1}^{m}|a_n| = t_m -t_k \to 0 .\tag{1}$$

I am really unsure about the above inequality

Well, try to prove it using $|a+b|\leq|a|+|b|$.

and how that has been used to prove the theorem ?

Keep the goal in mind: you want to show that $$|s_m-s_k|<\epsilon$$ for large enough $m$ and $k$.

Also I understand that |t_m -t_k| < e but I am unsure how this implies that t_m- t_k -> 0

$t_m-t_k\to 0$ is a bad writing. What it really means here is $|t_m-t_k|<\epsilon$ for "large enough" $m$ and $k$ where $\epsilon>0$ is assumed to be given.

Essentially the inequality in (1) tells you that $\{s_k\}$ is a Cauchy sequence.

Use Cauchy's criterion: the series is convergent if and only if for any $\varepsilon>0$, there exists a $N>0$ such that for any $m,k\ge N$, we have $\;\lvert s_m-s_k\rvert <\varepsilon$.

Now since the series is absolutely convergent, we do have a $N$ such that for any $m,k\ge N$, $\;\lvert t_m-t_k\rvert <\varepsilon$. By the triangle inequality, the same is a fortiori true for $\lvert s_m-s_k\rvert$.

Sketch of the proof of Cauchy's criterion:

It's the same criterion for sequences and for series. So let $(u_n)$ be a Cauchy sequence. Proving it has a limit goes through the following steps:

• We associate to $(u_n)$ two sequences: $(a_n)=(\inf\limits_{p\ge n}u_p)$ and $(b_n)=(\sup\limits_{p\ge n}u_p)$.
• Show $(a_n)$ is non-decreasing and $(b_n)$ is non-increasing. Furthermore, for all $n$, we have $$a_n\le u_n\le b_n.$$
• Deduce from Cauchy's property that $(b_n-a_n)$ tends to $0$, so they're adjacent sequences and converge to a common limit $L$. By the squeezing principle, $(u_n)$ also converges to $L$.
• I think you missed the point of the OP: he already has this proof. He just seems to be lost on the triangle inequaity. – DonAntonio Dec 17 '16 at 13:29
• @bernard I understand that |t_m -t_k| < e but I am unsure how this implies that t_m-t_k -> 0 – questiontime Dec 17 '16 at 13:32
• @iheartalgebraa: It is not that $t_m-t_k\to 0$ (how is it defined anyway? it's not a term of a sequence, as there are 2 indices), but that it can be made as small as you please, choosing $m,k$both large enough. This is simply Cauchy's criterion. – Bernard Dec 17 '16 at 13:37
• @Bernard can you expand on explaining the proof of cauchy's criterion because this is what is limiting my understanding of this proof. – questiontime Dec 17 '16 at 13:39
• I've added some details.Does that make it clearer to you? – Bernard Dec 17 '16 at 14:32

Since $\sum a_n$ is absolutely convergent, so $\sum |a_n|$ is convergent. And since $a_n\le|a_n|$, by comparison test, $\sum a_n$ is convergent.

• You can't use comparison test here as it is not given $\;a_n\;$ is positive.., in fact, if it is a positive series then the claim is very boring. The theorem is interesting only for non-positive series. – DonAntonio Dec 17 '16 at 13:25