Show that $\lim_{n\to\infty}(6n)^{\frac16}a_n=1$ with $(a_n)$ such that $\lim_{n\to\infty}a_n\sum_{j=1}^na_j^5=1$ 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.

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 $s_n^6 - s_{n-1}^6 \to 6$.
