# Proof of Riemann Rearrangement Theorem

I’m reading the proof of Riemann Rearrangement Theorem in T. Tao’s Analysis 1 textbook which can be found here
Rearrangement Thm
(the parts missing from the textbook, left as exercises for the reader, are completed by the user asking the question)
but I don’t understand the last line of the proof where the user says
“If $u_i <l_i$ then for all $u_i \leq k\leq l_i$ we therefore have…”;
to affirm that $S_k \to L$ shouldn’t one prove that it is always $S_{l_i}\leq S_{k}\leq S_{u_i}$ to be able to invoke the Squeeze Theorem?
I don’t understand how that proof accomplishes this.

Could someone explain this part of that proof or show me another way to finish the proof?

Best regards,

lorenzo.

#### Solutions Collecting From Web of "Proof of Riemann Rearrangement Theorem"

The paragraph just above what your are asking about completes the proof. The one you are asking about is not correct. The $u$s are the next to last in a run of negative terms we are adding, so they are the last partial sum that is greater than $L$ for a while. We add one more negative term and drop below $L$, so we start adding positive terms. Similarly, the $\ell$s are the next to last in a run of positive terms, so they are the last partial sum that is less than $L$ for a while. We add one more positive term and rise above $L$, so we start adding negative terms.

If $u_i \lt \ell_i$, $S_{u_i+1}$ is a local minimum, so for all $k$ such that $u_i+1 \le k \le \ell_i$ we would have $S_{u_i+1} \le S_k \le S_{\ell_i}\lt L \lt S_{u_i}$ Now since $|S_{u_i}-S_{u_i+1}|\to 0$ because it is one term and the terms are converging to zero we get the squeeze we want. The case of $\ell_i \lt u_i$ is similar.