Proof on Riemann's Theorem that any conditionally convergent series can be rearranged to yield a series which converges to any given sum.

I am looking at the proof of the following theorem from Apostol’s Mathematical Analysis. I am having trouble showing the last part that the author left to the reader.

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I’m trying to show that $y$ is the limit superior of this rearrangement. To do so, I need to show two things based on the definition of limit superior from the text. First, for every $\epsilon \gt 0$ there is an integer $N$ such that $n \gt N$ implies $h_n \lt y + \epsilon$.

Second, given $\epsilon \gt 0$ and $m \gt 0$, there is an integer $n \gt m$ such that $h_n \gt y-\epsilon$. Where I put $h_n$ as the rearrangement of the original series.

The second condition is immediately satisfied since for every $y_n$ there is a rearrangement greater than it by construction. However, I’m having trouble showing the first part. How can I guarantee that for any $\epsilon$, all but finitely many $h_n$ is less than $y+ \epsilon$. I don’t know how to show this part since our construction only guarantees that we have some rearrangement greater than every $y_n$.

Finally, how does this theorem lead to the conclusion that any conditionally convergent series of real terms can be rearranged to yield a series which converges to any prescribed sum?

I’d greatly appreciate it if anyone could rigorously establish the above facts for me.

Solutions Collecting From Web of "Proof on Riemann's Theorem that any conditionally convergent series can be rearranged to yield a series which converges to any given sum."

Since the original series $\sum a_{n}$ is convergent it follows that $p_{n}, q_{n}$ both tend to $0$.

Next note that $y_{n} \to y$ so that for any $\epsilon > 0$ you have a value $n_{0}$, such that $y_{n} < y + \epsilon/2$ for $n > n_{0}$. If you note carefully the selection of terms from sequences $p_{n}, q_{n}$ to make it just greater than $y_{n}$, you will notice that removing the last term of type $p_{n}$ will make the sum less than $y_{n}$. Since $p_{n} \to 0$, it follows that we will reach a point when this last term in the sum corresponding to $y_{n}$ is less than $\epsilon / 2$ and therefore the difference of this sum from $y_{n}$ is less than $\epsilon / 2$. It follows that the sum itself will be less than $y_{n} + \epsilon / 2$. Thus we can find the finite sums of type $\sum p – \sum q$ such that $$y_{n} < \sum p_{k_{i}} – \sum q_{k_{j}} < y_{n} + \epsilon / 2 < y + \epsilon $$ This is the way we find a value $n_{1}$ such that $t_{n} < y + \epsilon$ for all $n > n_{1}$.

If you want to make the rearranged series $\sum b_{n}$ to converge to a specific value $x$ simply choose $x = y$ in the construction mentioned in your post so that $\liminf\, t_{n} = \limsup\, t_{n} = \lim\, t_{n} = x$.