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Let $\{x_i^n\}_{i=0}^{m_n-1}$ a subdivision of $[0,1]$ s.t. $\max_{i=1,...,m_n-1}\Delta x_i^n\to 0$. Does $$\lim_{n\to \infty }\sum_{i=0}^{m_n-1}o(\Delta x_i^n)=0 \ \ ?$$

For me, it's always the case, but in my question here, I may have a doubt. The thing is I can't get a counter example (all example I have works). So can someone confirm if it's always true or provide a counter example ?

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  • $\begingroup$ I’m not sure I totally understand the notation. What do you mean when you write $o(\Delta x_i^n)$? $\endgroup$ – Clayton Aug 14 at 17:45
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I don't think so, if I understand your question.

If $\Delta x_i^n =\dfrac1{n} $ we can choose $o(\Delta x_i^n) =\dfrac1{n\ln(n)} $ and the sum of this diverges.

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  • $\begingroup$ Well, $\sum_{k=0}^{n-1}o(\Delta x_i^n)=\frac{1}{n\ln(n)}\sum_{i=0}^{n-1}=\frac{1}{\ln(n)}\to 0$, so it works... The thing is when $f(x+h)=f(x)+f'(x)h+o(h)$ I often see in text : $f(1)-f(0)=\sum_{i}(f(x_{i+1}^n)-f(x_i^n))=\sum_{i}f'(x_i^n)\Delta x_i^n+\sum_{i}o(\Delta x_i^n)\to \int_0^1 f'(x)dx$ because $\sum_{i}o(\Delta x_i^n)\to 0$ when $n\to \infty $. And this last limit is never rigorously justify in my lecture. $\endgroup$ – user657324 Aug 14 at 20:30
  • $\begingroup$ What do you mean by $\Delta x_i^n$? Is the exponent a power or an index? $\endgroup$ – marty cohen Aug 15 at 3:02
  • $\begingroup$ Sorry, no $\{x_i^n\}_{i=0}^{m_n-1}$ is a partition of $[0,1]$ s.t. $\max_{i=1,...,m_n-1}\Delta x_i^n\to 0$ when $n\to \infty $ where $\Delta x_i^n=x_{i+1}^n-x_i$. $\endgroup$ – user657324 Aug 15 at 7:21

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