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Let $(x_n)_{n \ge 1}$ defined as follows: $$x_1 \gt 0, x_1+x_2+\dots+x_n=\frac {1}{\sqrt {x_{n+1}}}.$$ Compute the limit $\lim _ {n \to \infty} n^2x_n^3.$

MY TRY: I thought about using Stolz-Cesaro lemma, but I couldn't get to an appropriate form that leads to an easier limit.

Martin Sleziak
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M. Stefan
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    Please use more descriptive titles. Otherwise many of your questions will just be titled "Sequence defined recursively". – Aryabhata Aug 06 '17 at 20:03

2 Answers2

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Let $y_n=x_1+\ldots+x_n$ for $n\ge1$. Then, the recursion reads $y^2_n\,(y_{n+1}-y_n)=1.$ We see that $y_{n+1}-y_n>0$ for $n\ge1.$ Since $y^2$ is monotone, it follows that $$n=\sum^n_{k=1}y^2_k\,(y_{k+1}-y_k)\le\int^{y_{n+1}}_{y_1}y^2\,dy=\frac13(y^3_{n+1}-y^3_1),$$ i.e. $$y_{n+1}\ge3^{1/3}n^{1/3}.\tag1$$ Consequently $$\frac1{y^2_n}\le3^{-2/3}\frac1{(n-1)^{2/3}},$$ and $$y_{n+1}=y_2+\sum^n_{k=2}\frac1{y^2_k}\le y_2+3^{-2/3}\sum^n_{k=2}\frac1{(k-1)^{2/3}}=y_2+3^{-2/3}\sum^{n-1}_{k=1}\frac1{k^{2/3}}.$$ Now we can estimate the RHS with a telescope: $$k^{1/3}-(k-1)^{1/3}=\frac1{k^{2/3}+k^{1/3}(k-1)^{1/3}+(k-1)^{2/3}}\ge\frac1{3k^{2/3}},$$ that means $$y_{n+1}\le y_2+3^{1/3}(n-1)^{1/3}.\tag2$$ (1) and (2) together give $$\lim_{n\rightarrow\infty}y_n\,n^{-1/3}=3^{1/3}.$$ Since $x_{n+1}=y_{n+1}-y_n=1/y^2_n,$ we have $$\lim_{n\rightarrow\infty}x_n\,n^{2/3}=3^{-2/3},$$ and thus $$\lim_{n\rightarrow\infty}x^3_n\,n^2=3^{-2}=\frac19.$$

  • I appreciate this answer a lot. How did you obtain the estimation $\frac1{3 k^\frac23} \le k^\frac13-(k-1)^\frac13$, if I might ask? (To estimate $\sum_{k=1}^{n-1} k^{-\frac23}$ I would have used an integral: $$1 + \sum_{k=2}^{n-1} k^{-\frac23} \le 1 + \int_2^n (x-1)^{-\frac23}, dx=3(n-1)^\frac13-2.)$$ – Giuseppe Negro Aug 07 '17 at 20:41
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    @Giuseppe Negro That's clear from $$k^{1/3}-(k-1)^{1/3}=\frac{k-(k-1)}{k^{2/3}+k^{1/3}(k-1)^{1/3}+(k-1)^{2/3}}\ge\frac1{3k^{2/3}},$$ since $k-1<k$. Of course, an integral does the job, too. –  Aug 08 '17 at 04:54
  • That's a clever trick. Thank you. – Giuseppe Negro Aug 08 '17 at 09:48
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This answer will use the Stolz-Cesaro lemma. $$L=\lim_{n \to \infty} n^2x_n^3$$ $$S_n=\sum_{k=1}^n{x_k} \quad\quad x_{n+1}=S_n^{-2}$$ We will try to find a recurrence relation that is easier to manipulate: $$\frac{x_{n+1}}{x_n}=\frac{S_n^{-2}}{S_{n-1}^{-2}}=\frac{S_{n-1}^{2}}{S_n^{2}}=\left (\frac{S_{n-1}}{S_n}\right)^2$$ $$S_{n-1}=S_n-x_n$$ $$\frac{x_{n+1}}{x_n}=\left(\frac{S_n-x_n}{S_n}\right)^2$$ $$\frac{x_{n+1}}{x_n}=\frac{1}{x_nS_n^2}=\left(\frac{1}{\sqrt{x_n}S_n}\right)^2$$ $$\left(\frac{S_n-x_n}{S_n}\right)^2=\left(\frac{1}{\sqrt{x_n}S_n}\right)^2$$ $$\frac{S_n-x_n}{S_n}=\frac{1}{\sqrt{x_n}S_n}$$ $$S_n-x_n=\frac{1}{\sqrt{x_n}}$$ $$S_n=x_n+\frac{1}{\sqrt{x_n}}=x_n+x_n^{-\frac{1}{2}}$$ $$x_{n+1}=\frac{1}{\left(x_n+x_n^{-\frac{1}{2}}\right)^2}$$ $$x_{n+1}=\frac{x_n}{\left(x_n^{\frac{3}{2}}+1\right)^2} \quad n>1$$ Because of the limit we want to evaluate, we make the substitution: $a_n=n^2x_n^3 \quad x_n=a_n^{\frac{1}{3}}n^{-\frac{2}{3}} \quad L=a_{\infty}$ $$a_{n+1}^{\frac{1}{3}}(n+1)^{-\frac{2}{3}}=\frac{a_n^{\frac{1}{3}}n^{-\frac{2}{3}}}{\left(\frac {\sqrt{a_n}}{n}+1\right)^2}$$ $$a_{n+1}(n+1)^{-2}=\frac{a_nn^{-2}}{\left(\frac {\sqrt{a_n}}{n}+1\right)^6}$$ $$a_{n+1}=\frac{a_n(n+1)^{2}}{n^2\left(\frac {\sqrt{a_n}}{n}+1\right)^6}$$ To get rid of the square root we make the substitution: $a_n=b_n^2 \quad b_n=\sqrt{a_n} \quad L=b_{\infty}^2$ $$b_{n+1}^2=\frac{b_n^2(n+1)^{2}}{n^2\left(\frac {b_n}{n}+1\right)^6}$$ $$b_{n+1}=\frac{b_n(n+1)}{n\left(\frac {b_n}{n}+1\right)^3}$$ $$\frac{b_{n+1}}{n+1}=\frac{b_n}{n\left(\frac {b_n}{n}+1\right)^3}$$ This next substitution is somewhat obvious: $c_n=\frac{b_n}{n} \quad b_n= nc_n \quad L=\lim_{n \to \infty} (nc_n)^2$ $$c_{n+1}=\frac{c_n}{\left(c_n+1\right)^3}$$ Because we want the limit in a usable form for the Stolz-Cesaro lemma, we make the substitution: $c_n=\frac{1}{d_n} \quad d_n=\frac{1}{c_n}$ $$L=\lim_{n \to \infty} \left(\frac{n}{d_n}\right)^2=\left( \lim_{n \to \infty} \frac{n}{d_n}\right)^2$$ $$L=\left( \lim_{n \to \infty} \frac{n+1-n}{d_{n+1}-d_n}\right)^2$$ $$L=\lim_{n \to \infty} \frac{1}{(d_{n+1}-d_n)^2}$$ $$\frac{1}{d_{n+1}}=\frac{1}{d_n\left(\frac{1}{d_n}+1\right)^3}$$ $$d_{n+1}=d_n\left(\frac{1}{d_n}+1\right)^3=\frac{(d_n+1)^3}{d_n^2}$$ $$d_{n+1}=\frac{d_n^3+3d_n^2+3d_n+1}{d_n^2}$$ $$d_{n+1}=d_n+3+\frac{3}{d_n}+\frac{1}{d_n^2}$$ $$\lim_{n \to \infty} \left(d_{n+1}=d_n+3+\frac{3}{d_n}+\frac{1}{d_n^2}\right)$$ $$\lim_{n \to \infty} \left(d_{n+1}=d_n+3\right)$$ $$\lim_{n \to \infty} d_{n+1}-d_n=3$$ $$L=\frac{1}{3^2}=\frac{1}{9}$$ All of this is assuming that $d_n$ diverges to $+\infty$. Since $x_1>0$, $d_1>0$ (substitutions considered) is the only case that needs to be analyzed. We will prove that $d_{n+1}>d_n$ for $d_n>0$. $$\frac{d_n^3+3d_n^2+3d_n+1}{d_n^2}>d_n$$ $$d_n^3+3d_n^2+3d_n+1>d_n^3$$ $$3d_n^2+3d_n+1>0$$ This is clearly true for $d_n>0$. Note that this also shows there are no solutions to $d_{n+1}=d_n$ for $d_n>0$. Therefore, we can conclude $L=\frac{1}{9}$

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