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Let $a_n$ be a sequence of real numbers such that $\lim(a_n\sum_{k = 1}^n a_k^2) = 1$ . Prove that $\lim((3n)^{\frac{1}{3}}a_n)=1$

I'm more concerned with how I can derive the prove of this question

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  • $\begingroup$ More concerned than what? $\endgroup$ – Clement C. Aug 19 '17 at 16:56
  • $\begingroup$ Also, as a tiny step towards a possible solution: can you show (1) $\lim_{n\to \infty}\sum_{k=1}^n a_k^2 = \infty$ and (2) $\lim_{n\to\infty} a_n =0$? Also, to give you some intuition (maybe): the statement is the discrete/series equivalent of the analogue statement in the real world: $\lim_{x\to\infty }\sqrt{A'(x)} A(x)=1$, or equivalently $\lim_{x\to\infty }A'(x) A(x)^2=1$ (can you see why?) So it should make "sense" that the solution is similar (in the discrete setting) to that of the differential equation $A'A^2 = 1$. $\endgroup$ – Clement C. Aug 19 '17 at 18:50
  • $\begingroup$ After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark ✓ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?. $\endgroup$ – Clement C. Aug 23 '17 at 11:50
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Let $S_n=a_1^2+\cdots+a_n^2$. The sequence $(S_n)_{n\ge1}$ is nondecreasing. If it is convergent then $\lim_{n\to\infty}a_n=0$ and this contradicts the hypotgesis $\lim_{n\to\infty}a_nS_n=1$. Thus, $\lim_{n\to\infty}S_n=+\infty$ and $\lim_{n\to\infty}a_n= \lim_{n\to\infty}\frac{a_nS_n}{S_n}=0$. Now $$S_{n}^3-S_{n-1}^3=S_{n}^3-(S_{n}-a_n^2)^3=3(a_nS_n)^2-3a_n^3(a_nS_n)+a_n^6$$ Hence $$\lim_{n\to\infty}(S_{n}^3-S_{n-1}^3)=3$$ So, by Cesàro's lemma we get $$\lim_{n\to\infty}\frac{S_{n}^3}{n}=3$$ Or equivalently $$\lim_{n\to\infty}\frac{\sqrt[3]{3n}}{S_{n}}=1$$ Finally $$\lim_{n\to\infty}\sqrt[3]{3n}a_n=\lim_{n\to\infty}(a_nS_n)\frac{\sqrt[3]{3n}}{S_{n}}=1$$ Done.

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  • $\begingroup$ Neat and concise! $\endgroup$ – Clement C. Aug 19 '17 at 18:57
  • $\begingroup$ @FelixKlein I took the liberty to build on your answer below -- let me know if the attribution sounds satisfactory. $\endgroup$ – Clement C. Aug 19 '17 at 19:27
  • $\begingroup$ Orgasmic proofs in MSE ladies and gentlemen...+1) $\endgroup$ – Marios Gretsas Aug 19 '17 at 20:54
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Generalizing Felix Klein's wonderful answer, I want to abstract a bit the statement and the argument:

Let $\alpha > 0$, and $(a_n)_n$ be a positive sequence such that $\lim_{n\to\infty} a_n\sum_{k=1}^n a_k^\alpha = 1$. Find an asymptotic equivalent for $a_n$.

(Note that the OP's question is for $\alpha=2$.)

  • First, letting $A_n \stackrel{\rm def}{=} \sum_{k=1}^n a_k^\alpha$ it is not hard to show that $\lim_{n\to\infty} A_n =\infty$, and $\lim_{n\to\infty} a_n =0$.

    (Indeed, $(A_n)_n$ converges in $[0,\infty]$ by monotone convergence; if its limit is finite, then this implies $\lim_{n\to\infty} a_n^\alpha =0$, and therefore $\lim_{n\to\infty} a_n =0$, and therefore $\lim_{n\to\infty} a_n A_n =0$ -- contradiction.)

  • Then, considering $S_n^{\alpha+1} - S_{n-1}^{\alpha+1}$, we have \begin{align} S_n^{\alpha+1} - S_{n-1}^{\alpha+1} &= S_n^{\alpha+1} - \left(S_n-a_n^\alpha\right)^{\alpha+1} = S_n^{\alpha+1} - S_n^{\alpha+1}\left(1-\frac{a_n^\alpha}{S_n}\right)^{\alpha+1} \\ &= S_n^{\alpha+1} - S_n^{\alpha+1}\left(1-(\alpha+1)\frac{a_n^\alpha}{S_n} + o\left(\frac{a_n^\alpha}{S_n}\right)\right) \\ &= S_n^{\alpha+1} - \left(S_n^{\alpha+1}-(\alpha+1)(a_n S_n)^{\alpha} + o\left((a_n S_n)^{\alpha}\right)\right) \\ &= (\alpha+1)(a_n S_n)^{\alpha} + o\left((a_n S_n)^{\alpha}\right) \\ &\operatorname*{\sim}_{n\to\infty} (\alpha+1)(a_n S_n)^{\alpha} \operatorname*{\sim}_{n\to\infty} \alpha+1 \end{align} where we used the fact (by the the first item we proved) that $\lim_{n\to\infty}\frac{a_n^\alpha}{S_n} = 0$ to perform a first-order Taylor expansion of $(1+u)^{\alpha+1}$ around $0$; and, for the last part, the fact that $\lim_{n\to\infty }(a_n S_n)^{\alpha} = 1$.

Now, why did we do that? This is a general technique: let's take a step back, and consider all our discrete sums and sequences as their integral and functional analogues. Then, we are considering $$ (S^{\alpha+1})' \approx S_n^{\alpha+1} - S_{n-1}^{\alpha+1} $$ and if we can show $\lim (S^{\alpha+1})' = c$ for some constant $c$, then by "integration" we will have $S^{\alpha+1}(x) \sim_{x\to \infty} cx$, i.e. $S(x) \sim_{x\to \infty} (cx)^{1/(\alpha+1)}$. That the handwavy intuition behind that step; let's make it formal.

  • Summing both sides, and using the fact that the series $\sum_{n}( S_n^{\alpha+1}-S_{n-1}^{\alpha+1})$ diverges as $\lim_{n\to\infty} S_n =\infty$ to argue that the LHS and RHS series will be asymptotically equivalent,* we get $$ S_n^{\alpha+1} - S_0^{\alpha+1} = \sum_{k=1}^n (S_k^{\alpha+1} - S_{k-1}^{\alpha+1}) \operatorname*{\sim}_{n\to\infty} (\alpha+1)n $$ i.e., as $S_0 = 0$, $ S_n^{\alpha+1} \operatorname*{\sim}_{n\to\infty} (\alpha+1)n. $

    Using one last time our assumption, we have $$ a_n \operatorname*{\sim}_{n\to\infty} \frac{1}{S_n} \operatorname*{\sim}_{n\to\infty} \boxed{\frac{1}{((\alpha+1)n)^{\frac{1}{\alpha+1}}}} $$ concluding the proof.

${}^*$ We can also rely on Cèsaro for this step.

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  • $\begingroup$ I would remove the absolute value and speak of positive sequences. $\endgroup$ – Felix Klein Aug 19 '17 at 19:41
  • $\begingroup$ @FelixKlein Since it doesn't change anything, why not. Done. $\endgroup$ – Clement C. Aug 19 '17 at 19:46

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