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.

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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.