Given a sequence $a_1,\ldots,a_n$, we can take the average $\bar{a}=\frac{\sum_{i=1}^n a_i}{n}$. For some real valued function $f$ it seems to me intuitively that $$\sum_{i=1}^n \frac{1}{f(a_i)}=\frac{n}{f(\bar{a})}.$$ Should this be true, I will go about trying to prove this via induction.
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1What does $$\lim_{n\rightarrow\infty}\sum_{i=1}^n \frac{1}{f(a_i)}=\frac{n}{f(\bar{a})}$$ mean? – Kavi Rama Murthy Jan 28 '22 at 04:59
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You are right - it is a bit nonsensical. I will edit. The question stands for the second equation though. – pestopasta Jan 28 '22 at 05:01
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1The second equation is obviously false for many nice functions and this has no connection with Law of Large Numbers. – Kavi Rama Murthy Jan 28 '22 at 05:04
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You are right about LLN -I was thinking about it wrong. Why is it obviously wrong? And is there some class of functions for which it holds? – pestopasta Jan 28 '22 at 05:07
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@pestopasta Let $g(x)=1/f(x)$ then $g(\bar a) = \frac{1}{n}\sum a_i$. If you require $g$ to be continuous and the equality to hold for all real sequences, the only solutions are linear functions $g(x) = mx +n$, see for example How to show that $f$ is a straight line if $f(\frac{x+y}{2})=\frac{f(x)+f(y)}{2}$? – dxiv Jan 28 '22 at 05:45
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More correct: If $f$ is continuos and non-zero in $[\min(\{a_k\}),\max(\{a_k\})]$ then exists $\bar{a}\in [\min(\{a_k\}),\max(\{a_k\})]$ (maybe not arithmetical mean) such that $\sum_{i=1}^n \frac{1}{f(a_i)}=\frac{n}{f(\bar{a})}$
Ivan Kaznacheyeu
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