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

Question

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

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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 $t_m-t_k \to 0$

Answer 1

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.

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

Answer 2

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.

Answer 3

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 $L$.