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My doubt is in second picture where author writes "If infinitely many $s_n$ are equal to $v$ ". Is it not true that this will always hold ? Because if finitely many terms of sequences are equal to $v$. Then after certain stage (past last dominating term) say Sup $ \{s_n : n > r_k\}$ < $v$. Thus contradicting claim made earlier that this set is constant sequence. Another point is where author claims that $s_{m_0} < v$ in equation 2. Is it not false ? Because $m_0$ is last dominating term and after that value of terms of sequence are less than $s_{m_0}$. Thus $v$ < $s_{m_0}$ Thanks

Sophie Clad
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1 Answers1

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  1. Consider the sequence defined by $s_n := - \frac{1}{n}$. Then $v_n = 0$ for all $n$, and yet finitely many (i.e. zero in this case) $s_n$s are equal to $v = 0$. The $v_n$s are not necessarily part of the sequence $(s_n)_n$, because the supremum of an infinite subset of the reals is not necessarily an element of said subset.

  2. The author is not claiming that $n_1$ is equal to $m_0$, only that $n_1 \geq m_0$, which is sufficient to proceed. You are partially correct, in that $v \leq s_{m_0}$, and so $m_0 \neq n_1$, but saying $n_1 \geq m_0$ does not guarantee that the case $n_1 = m_0$ will ever be true, just like you can rightfully write $1 \geq 0$ even though $1$ also satisfies $1 > 0$.

Bruno B
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  • Thanks. I am not sure about your first example. Because there dont seem to be finite dominating terms.. – Sophie Clad Dec 19 '24 at 23:02
  • @SophieClad You can add at the beginning of the sequence a finite number of $0$s if you'd like, which will act as dominant terms. Remember that "no term is dominant" is also a subcase of "finitely many terms are dominant", unless there's explicitly written that at least one term is dominant. I suppose the author here was slightly sloppy in dealing with the absence of dominating terms, but in that case just set $m_0$ to be the first index of your sequence and you'll be good to go. – Bruno B Dec 19 '24 at 23:11
  • also $v =s_{m_0} $ doesnot seem to be correct because individual terms of sequence $v_N$ are all less than last dominating term so how can limit equal to that ? – Sophie Clad Dec 19 '24 at 23:13
  • @SophieClad The counterexample I provided, to which we add a $0$ as first term $s_0$ to have a dominant term, satisfies $v = s_0 = 0$. – Bruno B Dec 19 '24 at 23:16
  • hi....Can you please explain me 3rd and 4th paragraphs of last answer(by Isky Mathews) in this question.. https://math.stackexchange.com/questions/4249348/interchange-summation-indexes-m%c3%b6bius-inversion-formula?rq=1 – Sophie Clad Dec 22 '24 at 01:01