1
$\begingroup$

I'm in a process of proving the last part of Riemann's Theorem on conditionally convergent series.

The theorem states: Let $\sum a_n$ be a conditionally convergent series with real-valued terms. Let $x$ and $y$ be given numbers in the closed interval [$-\infty, +\infty$], with $x\le y$. Then there exists a rearrangement $\sum b_n$ of $\sum a_n$ such that $\limsup S_k=y$ and $\liminf S_k=x$, where $S_k = b_1+...+b_n$.

Here is a sketch of my proof:

Since $\sum a_n$ is conditionally convergent, I get that that the derived series $\sum a^+_n$ and $\sum a^{^{-}}_n$ are both diverge sequence. Where $\sum a^+_n$ denotes the sum of positive terms of $\sum a_n$; $\sum a^{^{-}}_n$ denotes the sum of negative terms of $\sum a_n$.

Construct two sequences of real numbers, say {$x_n$} and {$y_n$} such that $\lim_{n\to\infty} x_n=x$, $\lim_{n\to\infty} y_n=y$ with $x_n\le y_n$, $y_1\gt 0$

Now, to begin, add, in order, just enough of the first positive terms of $\sum a_n$ so that their sum is greater than $y_1$. This is possible by the observation. Then we will need $M_1$ terms of $\sum a_n$ to do this. Therefore, the first partial sum of our rearrangment is $S_{M_{1}}=\sum_{n=1}^{M_1} a_{\sigma(n)}>y_1$.

Now, to $S_{M1}$, add just enough negative terms from the original series so that the sum is less than $x_1$. We will need a total of $N_1$ terms of the original series to do this. Therefore, the first partial sum here is $S_{N_{1}}=\sum_{n=1}^{N_1} a_{\sigma(n)}<x_1$.

Now proceed inductively: having chosen $M_k$ and $N_k$ such that $S_{M_{k}}=\sum_{n=1}^{M_k} a_{\sigma(n)}>y_k$ and $S_{N_{k}}=\sum_{n=1}^{N_k} a_{\sigma(n)}<x_k,\ $ to $S_{N_{k}}$ we add just enough positive terms from the original series, so that $S_{M_{k+1}}=\sum_{n=1}^{M_{k+1}} a_{\sigma(n)}>y_{k+1}$, and to this series just enough negative terms so that $S_{N_{k+1}}=\sum_{n=1}^{N_{k+1}} a_{\sigma(n)}<x_{k+1}$.

By construction, the sequence of partial sums $S_k=\left \{ S_{M_1},S_{N_1},\cdots , S_{M_k},S_{N_k}\right \}_{k\geq 1}$ should be a rearrangement of the original series such that $\limsup S_k=y$ and $\liminf S_k=x$.

But I have no idea how to show $\limsup_{k\to\infty} S_k=y$ and $\liminf_{k\to\infty} S_k=x$, could someone provide a proof of this please? Thanks!

$\endgroup$
  • $\begingroup$ You are missing a key observation. Because the original series converges, albeit conditionally, you know that $a_n\to 0$. If you don't use this then your proof will get no further than $\limsup S_k \geq y$ and $\liminf S_k \leq x$. $\endgroup$ – B. S. Thomson Jan 31 '16 at 21:55
  • $\begingroup$ Thanks, but I'm still stuck on deriving $\limsup S_k=y$ and $\liminf S_k=x$, could you give a precise proof of this please? $\endgroup$ – jax8901 Jan 31 '16 at 22:07
  • $\begingroup$ This is just textbook, but (as you no doubt know) many textbooks prove only the simpler cases. I found this one books.google.com/books?isbn=0817682929 and the relevant pages do a rather full discussion of the theorem and Google even lets you see those pages (but not all of the text). Good luck. $\endgroup$ – B. S. Thomson Jan 31 '16 at 23:43

Your Answer

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

Browse other questions tagged or ask your own question.