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Problem:Let postive real sequence$\{a_{n}\}$ satisfy $\displaystyle\lim_{n\to\infty}a_{n}\left(\sum_{i=1}^{n}a_{i}^{p}\right)=1$,where $p>-1$,Find the limit. $$ \lim_{n\to\infty}\frac{n}{\ln{n}}\left(\frac{1}{p+1}-na_{n}^{p+1}\right) $$ Here is my approach:

First let estimate $a_{n}^{p+1}$,Let us note $S_{n}=\sum_{i=1}^{n}a_{i}^{p}$,it is easy to check $$\lim_{n\to\infty}{a_{n}}=0,\lim_{n\to\infty}S_{n}=+\infty$$ and by O.Stolz Theorem we have $$\lim_{n\to\infty}na_{n}^{p+1}=\lim_{n\to\infty}a_{n}^{p+1}S_{n}^{p+1}\cdot\lim_{n\to\infty}\frac{n}{S_{n}^{p+1}}=\lim_{n\to\infty}\frac{1}{S_{n+1}^{p+1}-S_{n}^{p+1}} $$ but \begin{align*} S_{n+1}^{p+1}-S_{n}^{p+1}&=S_{n+1}^{p+1}-(S_{n+1}-a_{n+1}^{p})^{p+1}\\ &=S_{n+1}^{p+1}-\sum_{k=0}^{p+1}C_{p+1}^{k}(-1)^{k}S_{n+1}^{p+1-k}\cdot a_{n+1}^{pk}\\ &=(p+1)S_{n+1}^{p}a_{n+1}^{p}-\frac{(p+1)p}{2!}S_{n+1}^{p-1}a_{n+1}^{2p}+o(S_{n+1}^{p-1}a_{n+1}^{2p})\\ &=(p+1)+o(1) \qquad (n\to+\infty) \end{align*} Therefore $$(p+1)a_{n}^{p+1}\sim \frac{1}{n} $$ to be convince note $A=\frac{1}{p+1}$, \begin{align*} &\lim_{n\to\infty}\frac{n}{\ln{n}}\left(A-na_{n}^{p+1}\right)\\ &=\lim_{n\to\infty}\frac{AS_{n}^{p+1}-na_{n}^{p+1}S_{n}^{p+1}}{\ln{n}}\cdot\lim_{n\to\infty}\frac{na_{n}^{p+1}}{S_{n}^{p+1}a_{n}^{p+1}}\\ &=\frac{1}{p+1}\lim_{n\to\infty}\frac{AS_{n}^{p+1}-na_{n}^{p+1}S_{n}^{p+1}}{\ln{n}}\\ &=\frac{1}{p+1}\lim_{n\to\infty}\frac{A(S_{n+1}^{p+1}-S_{n}^{p+1})-[(n+1)a_{n+1}^{p+1}-na_{n}^{p+1}S_{n}^{p+1}]}{\ln\left(1+\frac{1}{n}\right)}\\ &=\frac{1}{p+1}\lim_{n\to\infty}n\left[A\left(S_{n+1}^{p+1}-(S_{n+1}-a_{n+1}^{p})^{p+1}\right)-((n+1)a_{n+1}^{p+1}S_{n+1}^{p+1}-na_{n}^{p+1}S_{n}^{p+1}) \right]\\ &=\frac{1}{p+1}\lim_{n\to\infty}n\left[\frac{1}{p+1}\left((p+1)S_{n+1}^{p}a_{n+1}^{p}-\frac{(p+1)p}{2}S_{n+1}^{p-1}a_{n+1}^{2p}+o(S_{n+1}^{p-1}a_{n+1}^{2p}) \right)-((n+1)a_{n+1}^{p+1}S_{n+1}^{p+1}-na_{n}^{p+1}S_{n}^{p+1})\right]\\ &=\frac{1}{p+1}\lim_{n\to\infty}n\left[S_{n+1}^{p}a_{n+1}^{p}-(n+1)S_{n+1}^{p+1}a_{n+1}^{p+1}+na_{n}^{p+1}S_{n}^{p+1}-\frac{p}{2}S_{n+1}^{p-1}a_{n+1}^{2p} \right] \end{align*} I want to prove $$ \mathop {\lim }\limits_{n \to \infty } n\left[ {S_{n + 1}^pa_{n + 1}^p - \left( {n + 1} \right)S_{n + 1}^{p + 1}a_{n + 1}^{p + 1} + na_n^{p + 1}S_n^{p + 1}} \right] = 0$$ But I stuck here.Can someone help me? Thank you very much!

pxchg1200
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  • $$ \mathop {\lim }\limits_{n \to \infty } \frac{n}{{\ln n}}\left( {\frac{1}{{p + 1}} - na_n^{p + 1} } \right) = \frac{1}{{p + 1}}\underbrace {\mathop {\lim }\limits_{n \to \infty } \frac{n}{{\ln n}}}{\mathop \downarrow \limits\infty } - \mathop {\lim }\limits_{n \to \infty } \frac{n}{{\ln n}}na_n^{p + 1} $$ – Mohammad W. Alomari Jul 05 '14 at 15:16

1 Answers1

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The limit does not exist. The given conditions can only ensure $\lim_{n\to\infty}na_{n}^{p+1}=\frac{1}{p+1}$, but not able to dictate the rate of the convergence.

To make it simple, let $p=0$, we have $\lim_{n\to\infty}na_{n}=1$, so $a_n=\frac{1}{n}+o(\frac{1}{n})$. We give 2 examples:

  1. $a_n=\frac{1}{n}$

$\lim_{n\to\infty}\frac{n}{\ln{n}}\left(\frac{1}{p+1}-na_{n}^{p+1}\right)=\lim_{n\to\infty}\frac{n}{\ln{n}}\left(1-n\frac{1}{n}\right)=0$

  1. $a_n=\frac{1}{n}+\frac{1}{n\ln{n}}$

$\lim_{n\to\infty}\frac{n}{\ln{n}}\left(\frac{1}{p+1}-na_{n}^{p+1}\right)$

$=\lim_{n\to\infty}\frac{n}{\ln{n}}\left(-\frac{1}{\ln{n}}\right)=-\infty$

lion
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