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Let $p\ge -1$ and $\{a_n\}$ be a sequence of positive reals satisfaying $\displaystyle \lim_{n \to + \infty} a_n \left( \sum_{k=1}^n a_k^p \right) =1$
Find the limit $$\lim_{n \to \infty} \frac{n}{\log n} \left( \frac{1}{p+1} -na_n^{p+1}\right)$$

I am not even sure that this sequence converges, I mean I don't know how to start. Can you give me a hint ?

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    $\begingroup$ Replace the sum by an integral and solve the continuous problem instead $\endgroup$
    – M.LTA
    Oct 18 '15 at 10:14
  • $\begingroup$ How am I supposed to do that, integral methods can not work here ? $\endgroup$
    – user279923
    Oct 18 '15 at 10:22
  • $\begingroup$ I'll write the begining $\endgroup$
    – M.LTA
    Oct 18 '15 at 10:36
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Case where $p \geqslant 0$ : First, let $\displaystyle S_n=\sum_{k=1}^n a_k^p$. If $\sum_{S_n}$ converges we have $a_n \to 0$ which is impossible. Then we have $S_n \to \infty$ and $a_n \to 0$.
Note that $$S_{n}-S_{n-1}=a_n^p \sim \frac{1}{S_n^p}$$ The continuous analogy is $y'=\frac{1}{y^p}$ wich is $(y^{p+1})'=p+1$ This show you to look at : $$S_{n+1}^{p+1}-S_n^{p+1}=S_n^{p+1} \left( \left( 1+ \frac{a_{n+1}^p}{S_n} \right)^{p+1} -1 \right) =(p+1)S_{n}^{p+1}\frac{a_{n+1}^p}{S_n} + o(1)$$ But you can see that $1-\frac{S_n}{S_{n+1}}=\frac{a_{n+1}^p}{S_{n+1}} \to 0$ so we have : $$S_{n+1}^{p+1}-S_n^{p+1} = p+1 +o(1)$$ and $S_n \sim (p+1)n$ thus : $$na_n^{p+1} \sim \frac{1}{p+1} $$ This explain what your question is. Then we set $y_n:=S_{n}^{p+1}-(p+1)n$ and you get : $$y_{n+1}-y_n \sim \frac{p}{2n}$$ so that $y_n \sim \frac{p \log n}{2}$ and thus you can conclude and find that the limit is $\displaystyle \frac{p}{(p+1)^2}$

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  • $\begingroup$ Nice method, I will work on it $\endgroup$
    – user279923
    Oct 18 '15 at 12:27

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