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Suppose that $p_k>0, k=1,2,\dots, $ and $$ \lim_{n\to\infty} \frac{p_{n}}{p_1+p_2+\dots+p_n}=0,\quad \lim_{n\to\infty} a_n=a. $$

Show that $$ \lim_{n\to\infty}\frac{p_1a_n+p_2a_{n-1}+\dots+p_na_1}{p_1+p_2+\dots+p_n}=a $$

The hints are much appreciated. I don't want complete proof.

Thanks for your help.

Mhenni Benghorbal
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Paul
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1 Answers1

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Hint: at some $N$, $a_N$ is close enough to $a$ (how close is that?) and so is $a_n$ for $n>N$. For such an $n$, we can say that the sum is at least equal to $$ \frac{p_1a_N+p_2a_{N}+\dots+p_{n-N}a_N}{p_1+p_2+\dots+p_n} + \frac{p_{n-N+1}a_{N-1}+\dots+p_na_1}{p_1+p_2+\dots+p_n} $$ How can we show that the term on the left tends to $a_N$ while the term on the right tends to $0$?

Ben Grossmann
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  • Could you kindly tell me why does the term on the right side tend to 0? – Paul Jun 12 '13 at 01:34
  • Note that the term on the right will always have $N-1$ terms in the numerator, and an increasing number of terms in the denominator. Using the first fact, we can show that $$\lim_{n\to\infty} \frac{a_1p_{n}+a_2p_{n-1}}{p_1+p_2+\dots+p_n}=0$$ (how?). By that same logic, we can extend that for any finite number of terms in the numerator. – Ben Grossmann Jun 12 '13 at 01:45