# 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.

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 $\liminf\, t_{n} = \limsup\, t_{n} = \lim\, t_{n} = x$.