# If the series is absolutely convergent, then it is also conditionally convergent.

Let $$\sum_{n=m}^{\infty}a_{n}$$ be a formal series of real numbers. If the series is absolutely convergent, then it is also conditionally convergent. Furthermore, in this case we have the triangle inequality:

\begin{align*} \left|\sum_{n=m}^{\infty}a_{n}\right| \leq \sum_{n=m}^{\infty}|a_{n}| \end{align*}

MY ATTEMPT (EDIT)

Since $$S_{n} = |a_{m}| + |a_{m+1}| + \ldots + |a_{n}|$$ converges, it is a Cauchy sequence.

Therefore for every $$\varepsilon > 0$$, there exists a natural number $$N\geq m$$, such that \begin{align*} p \geq q\geq N \Longrightarrow |a_{p} + a_{p-1} + \ldots + a_{p-q+1}| \leq |S_{p} - S_{q}| \leq \varepsilon \end{align*} whence we conclude that $$s_{n} = a_{m} + a_{m+1} + \ldots + a_{n}$$ converges, because it is Cauchy too.

Similar reasoning proves that $$|a_{m} + a_{m+1} + \ldots + a_{N}|$$ converges.

This is because $$||x|-|y|| \leq |x - y|$$.

Consequently, one has that

\begin{align*} \left|\sum_{n=m}^{N}a_{n}\right| \leq \sum_{n=m}^{N}|a_{n}| \Longrightarrow \lim_{N\rightarrow\infty}\left|\sum_{n=m}^{N}a_{n}\right| \leq \lim_{N\rightarrow\infty} \sum_{n=m}^{N}|a_{n}| \Longrightarrow \left|\sum_{n=m}^{\infty}a_{n}\right| \leq \sum_{n=m}^{\infty}|a_{n}| \end{align*} and we are done.

Could someone please double-check my arguments?

• The result you applied assumes existence of $\sum a_n$. You did not prove that this sum exists. Use Cauchy criterion. – Kavi Rama Murthy Apr 21 at 23:52
• $c_n \leq d_n$ implies that $\lim c_n \leq \lim d_n$, if $\lim c_n$ and $\lim d_n$ exist. – Matthew Leingang Apr 21 at 23:56
• Thanks @KaviRamaMurthy for the comment. Could you please check if I am reasoning correctly after the edit? – BrickByBrick Apr 22 at 0:12
• Yes, you have now proved convergence. – Kavi Rama Murthy Apr 22 at 0:16
• Thanks @MatthewLeingang for the comment. I have fixed it. Could you please check my new answer? – BrickByBrick Apr 22 at 0:34