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Aug 16, 2023 at 16:33 vote accept rims
Aug 16, 2023 at 14:46 comment added Iosif Pinelis We have $p_j^n (x_j^n)^2\le V$ for all $j$ and $|x_j^n|\to\infty$ for $j\in J^c$. So, for $j\in J^c$ we have $|p_j^n x_j^n|=\dfrac{p_j^n (x_j^n)^2}{|x_j^n|}\le \dfrac V{|x_j^n|}\to0$. So, $p_j^n x_j^n\to0$ for $j\in J^c$. Are any other details needed?
Aug 16, 2023 at 1:47 comment added rims Yes, when you say for each $j \in J^c$, we have $p_j^* = 0$ and $p_j^n x_j^n \to 0$, how does $p_j^n x_j^n \to 0$ follow? Like what if $p_j^n \sim 1/n$ and $x_j^n \sim n^2$ ?
Aug 16, 2023 at 1:02 comment added Iosif Pinelis There a number of ways to see that any sequence in $\mathbb R$ has a subsequence converging to a point in $[-\infty,\infty]$. One is as follows: If the sequence is bounded from above and from below, use the Bolzano–Weierstrass theorem. If the sequence is not bounded from above, then it has a subsequence converging to $\infty$. If the sequence is not bounded from below, then it has a subsequence converging to $-\infty$. So, in each of the three cases, the sequence has a subsequence converging to a point in $[-\infty,\infty]$. Are any other details needed?
Aug 15, 2023 at 23:34 comment added rims Thank you for your response. When you are passing to convergent subsequences, is that a consequence of Bolzano–Weierstrass theorem for extended reals? I have only heard of the theorem for a bounded subset of reals so getting a convergent subsequence for $(p_1^n, p_2^n, p_3^n)$ makes sense but how do you get a convergent subsequence for $(x_1^n, x_2^n, x_3^n)$?
Aug 15, 2023 at 21:18 comment added Iosif Pinelis Do you have a response to the answer below?
Aug 15, 2023 at 2:09 answer added Iosif Pinelis timeline score: 2
Aug 14, 2023 at 15:19 review Close votes
Aug 19, 2023 at 3:02
S Aug 14, 2023 at 14:48 review First questions
Aug 14, 2023 at 18:34
S Aug 14, 2023 at 14:48 history asked rims CC BY-SA 4.0