0
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.