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Can this ever be the case? $$\lim\limits_{N\to\infty}\frac{\sum\limits_{i=1}^N a_i}{\sum\limits_{i=1}^N b_i} = \lim\limits_{N\to\infty}\frac{1}{N}\sum_{i=1}^N\frac{a_i}{b_i}$$ with $a_i>0$, $b_i>0$, $a_i<b_i$. As others pointed out simulations indicate convergence, but is there formal ground to it?

Community
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user330617
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  • @Crostul I have updated the question, giving more details. – user330617 Apr 12 '16 at 11:25
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    Your question doesn't make sense. $\frac{1}{N}\sum_{i=1}^N\frac{a_i}{b_i}$ depends on $N$. Do you actually mean $\lim\limits_{N\to\infty}\frac{\sum\limits_{i=1}^N a_i}{\sum\limits_{i=1}^N b_i} = \lim\limits_{N\to\infty} \frac{1}{N}\sum_{i=1}^N\frac{a_i}{b_i} $? – Gabriel Romon Apr 12 '16 at 11:28
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    @LeGrandDODOM Yes, thanks for pointing that out – user330617 Apr 12 '16 at 11:30

3 Answers3

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For a simple counterexample, take $a_i=\dfrac1{3^i}$ and $b_i=\dfrac1{2^i}$.

Then the LHS ratio tends to a finite value ($\frac12$), while the RHS tends to a finite value ($2$) over $N$, i.e. $0$.

  • Could you point out where I made error in my answer? I have tried to prove it as an exercise so it would help me to know what have I done wrong. – student Apr 20 '16 at 19:56
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Hint: try using Stolz theorem.

If the limit $\lim_{n\to\infty} \frac{a_n}{b_n}$ exists and is equal to $L$, then your limit is as well equal to $L$.

lisyarus
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user26977
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  • I might be missing something but if $a_n$, $b_n$ are sums over first $n$ elements, then $\lim_{n\to\infty}\frac{a_n}{b_n}$ is indeed the same - by definition. – user330617 Apr 12 '16 at 15:25
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We want to show \begin{equation} \lim\limits_{n\to\infty}\frac{\sum\limits_{i=1}^{n}{a_i}}{\sum\limits_{i=1}^{n}{b_i}}=\lim\limits_{n\to\infty}\frac{1}{n}\sum\limits_{i=1}^{n}\frac{a_i}{b_i} \end{equation} Assuming right hand side is convergent, take \begin{equation} \lim\limits_{n\to\infty}\frac{1}{n}\sum\limits_{i=1}^{n}\frac{a_i}{b_i}\to\bar{\varepsilon} \end{equation} rewrite as \begin{equation} \frac{\sum\limits_{i=1}^{\infty}{a_i}}{\sum\limits_{i=1}^{\infty}{b_i}}=\bar\varepsilon \end{equation} let \begin{equation} \varepsilon_i = \frac{a_i}{b_i} \end{equation} multiply right hand side by 1 and substitute $\varepsilon_i$ \begin{equation} \frac{\sum\limits_{i=1}^{\infty}{\varepsilon_ib_i}}{\sum\limits_{i=1}^{\infty}{b_i}}=\frac{\sum\limits_{i=1}^{\infty}\bar\varepsilon{b_i}}{\sum\limits_{i=1}^{\infty}{b_i}} \end{equation} substitute $\varepsilon_i=\bar\varepsilon+\Delta\varepsilon_i$ \begin{equation} \frac{\sum\limits_{i=1}^{\infty}{\bar\varepsilon b_i}+\sum\limits_{i=1}^{\infty}{\Delta\varepsilon_i b_i}}{\sum\limits_{i=1}^{\infty}{b_i}}=\frac{\sum\limits_{i=1}^{\infty}\bar\varepsilon{b_i}}{\sum\limits_{i=1}^{\infty}{b_i}} \end{equation} Then it suffices to show \begin{equation} \frac{\sum\limits_{i=1}^{\infty}{\Delta\varepsilon_i b_i}}{\sum\limits_{i=1}^{\infty}{b_i}}\to 0 \end{equation} Since $\forall i:a_i>0, b_i>0$ we obtain inequality \begin{equation} \frac{\sum\limits_{i=1}^{\infty}{\Delta\varepsilon_i b_i}}{\sum\limits_{i=1}^{\infty}{b_i}}<\frac{\left(\sum\limits_{i=1}^{\infty}{\Delta\varepsilon_i}\right)\left(\sum\limits_{i=1}^{\infty}{b_i}\right)}{\sum\limits_{i=1}^{\infty}{b_i}} \end{equation} \begin{equation} \frac{\sum\limits_{i=1}^{\infty}{\Delta\varepsilon_i b_i}}{\sum\limits_{i=1}^{\infty}{b_i}}<{\sum\limits_{i=1}^{\infty}{\Delta\varepsilon_i}} \end{equation} where \begin{equation} {\sum\limits_{i=1}^{\infty}{\Delta\varepsilon_i}}\to 0 \end{equation} due to expectation $\mathrm{E}[\Delta\varepsilon_i]=0$ by its definition

student
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