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Let $a_1, a_2, ...$ be a sequence of positive numbers such that $\frac{\sum_{j = 1} ^ n a_j}{n} \to \infty$ as $n \to \infty$. What can we say about the behavior of $$\frac{(\sum_{j = 1} ^ n a_j)^p}{n^{p - 1} \sum_{j = 1} ^ n a_j ^ p}$$ as $n \to \infty$ for $p > 1$? If I have my way, it should go to $0$ under these conditions, but I haven't been able to show this, I suspect because I'm missing some inequality. I can use Jensen's inequality to show that the expression above is in $[0, 1]$, but I guess I need something sharper to get it to go to $0$; maybe some inequality that leaves a factor of $\sum a_j / n$ in the denominator to exploit.

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Not neccesarily zero: Let $a_{j}=j$. Then $$\sum_{j\leq n}a_{j}\sim\frac{n^{2}}{2}\ \text{and}\ \sum_{j\leq n}a_{j}^{p}\sim\frac{1}{p+1}n^{p+1}$$so that $$\frac{\left(\sum_{j\leq n}a_{j}\right)^{p}}{n^{p-1}\sum_{j\leq n}a_{j}^{p}}\sim\frac{p+1}{2^{p}}.$$

Hölder Mean: We can write the terms in a slightly nicer way. Notice $$\frac{\left(\sum_{j\leq n}a_{j}\right)^{p}}{n^{p-1}\sum_{j\leq n}a_{j}^{p}}=\frac{\left(\frac{\sum_{j\leq n}a_{j}}{n}\right)^{p}}{\left(\frac{\sum_{j\leq n}a_{j}^{p}}{n}\right)}=\left(\frac{M_{1}(a_{1},\dots,a_{n})}{M_{p}(a_{1},\dots,a_{n})}\right)^{p}$$ where $M_{r}(\boldsymbol{x})$ is the generalized Hölder mean. For $p>1$ we know that $M_{1}(\boldsymbol{x})\leq M_{p}(\boldsymbol{x})$ so that the above quantity is always in $[0,1]$.

Sequence going to zero: We cannot say too much more, such as a lower bound, as it is possible to construct a sequence which goes to zero.

Let $a_{j}=e^{j}$. Then $$\sum_{j\leq n}a_{j}\sim e^{n}\ \text{and}\ \sum_{k\leq n}a_{j}^{p}\sim\frac{1}{p}e^{np}.$$ Consequently $$\frac{\left(\sum_{j\leq n}a_{j}\right)^{p}}{n^{p-1}\sum_{j\leq n}a_{j}^{p}}\sim\frac{\left(e^{n}\right)^{p}}{n^{p-1}\left(\frac{1}{p}e^{np}\right)}\sim\frac{p}{n^{p-1}}\rightarrow0.$$

Hope that helps,

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    $\begingroup$ Also perhaps of interest: $a_j = j^k$ for $k > 0$ gives $ \frac{(\sum_{j\le n} a_j)^p}{n^{p-1} \sum_{j\le n} a_j^p} \sim \frac{kp+1}{(k+1)^p}$ which could be anything in $(0,1)$. $\endgroup$ – Robert Israel Aug 7 '11 at 20:32
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    $\begingroup$ And, I think, $a_j = \ln j$ gives a limit of 1. $\endgroup$ – Robert Israel Aug 7 '11 at 20:39
  • $\begingroup$ @Robert Israel: Thanks for pointing that out! $\endgroup$ – Eric Naslund Aug 7 '11 at 20:58
  • $\begingroup$ Meh, this makes me unhappy since I was hoping to use it to prove an analogous statement about random variables that I know is true. The analogous statement is that if you replace the sequence $a_n$ by a sequence of iid positive random variables such that $X_n \notin L_1$ then the limit is $0$ almost surely. I guess I actually have to use probability theory in order to solve it :( $\endgroup$ – guy Aug 7 '11 at 21:09
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    $\begingroup$ @guy, the reason why your analogous statement for non integrable iid random variables holds is indeed truly stochastic, namely it is the fact that $X_1+\cdots+X_n$ is of the order of the maximum $M_n=\max\{X_1,\ldots,X_n\}$ up to time $n$. Similarly, $X_1^p+\cdots+X_n^p$ is of the order of $M_n^p$ hence the ratio you consider should converge almost surely to zero because of the power of $n$ in the denominator. $\endgroup$ – Did Aug 8 '11 at 10:19

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