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Show that $$ \lim_{n\to\infty}(6n)^{\frac16}a_n=1, $$ where $(a_n)$ is a sequence of nonnegative real numbers such that $\lim_{n\to\infty}a_n\sum_{j=1}^na_j^5=1.$

I recently got stuck on this problem. I can't seem to make an approach because this is the first I've seen this sort of problem.

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  • $\begingroup$ Very interesting question! My first thought was to do a case analysis on the convergences of $a_n$ and $b_n:=\sum a_j^5$. By this you can see that $b_n$ cannot be convergent, because otherwise $a_n\to0$, and as the limits of both $a_n$ and $b_n$ exist, we could separate the limit and see it vanishes, which is a contradiction. $\endgroup$ – Alp Uzman Mar 26 '15 at 17:23
  • $\begingroup$ So what we have is that $b_n\to\infty$, and by hypothesis the "speed" of $a_n$ vanishing keeps up with the "speed" of $b_n$ increasing unboundedly. We need to show that the "speed" of $a_n$ actually matches to that of $(6n)^{1/6}$. $\endgroup$ – Alp Uzman Mar 26 '15 at 17:26
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Proof Let $s_n = \sum_{k=1}^na_k^5$. We know that $$a_ns_n \to 1\text{ as }n\to\infty.$$ We want to show $$(6n)^{1/6}a_n\to 1\text{ as }n\to\infty.$$ It is enough to show $$\frac{s_n}{(6n)^{1/6}}\to 1 \text{ or equivalently }\frac{s_n^6}{n}\to 6\text{ as }n\to\infty.$$

Stolz–Cesàro theorem states:

Let $a_n$ and $b_n$ be two sequences, with $b_n$ unbounded and increasing. Then,

$$\lim_{n\to\infty}\frac{a_n}{b_n}=\lim_{n\to\infty}\frac{a_{n+1}-a_n}{b_{n+1}-b_n}$$

if the second limit exists.

Therefore it suffices to show $$s_n^6-s_{n-1}^6\to 6 \text{ as }n\to\infty.$$ Note that $$s_n^6-s_{n-1}^6 = s_n^6 - (s_n-a_n^5)^6 = 6a_n^5s_n^5 - 15a_n^{10}s_n^4 + 20a_n^{15}s_n^3 -15a_n^{20}s_n^2 + 6a_n^{25}s_n-a_n^{30}.$$ As $s_n$ is a non-decreasing sequence, if the sequence is bounded, then $\lim s_n= s$, and so $0 = \lim s_{n}-s_{n-1}=\lim a_n^5 = 1/s^5$, which is impossible. Therefore $s_n$ is unbounded and so $a_n^i s_n^j\to 0$ if $i > j$. Therefore $s_n^6 - s_{n-1}^6 \to 6$.

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  • $\begingroup$ The motivation of the solution is through the continuous analog of the question. One may find that the solution to the continuous variant needs L'Hôpital's rule. In the discrete version, Stolz-Cesàro theorem is the counterpart of L'Hôpital's rule. $\endgroup$ – Zilin J. Mar 26 '15 at 18:07

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