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*}


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?

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

Most definitions I know of conditional convergence states (e.g. https://en.wikipedia.org/wiki/Conditional_convergence)

A series is conditionally convergent if it is convergent but not absolutely convergent.

Which means also that if a series is absolutely convergent, it cannot be the case that it is also conditionally convergent.

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  • $\begingroup$ Indeed, you are right. But the author of the book which I am using to study says "we consider the class of conditionally convergent series to include the class of absolutely convergent series as a subclass" $\endgroup$ – BrickByBrick Apr 22 at 0:08

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