Modification on a sequence from VJIMC

The first two limits of the following problem were proposed at VJIMC, 2005, Category I.

The exact value of the last limit was proposed by a user at https://artofproblemsolving.com , where you can also see his solution along other lines.

 Let $(x_n)_{n\ge 2}$ be a sequence of real numbers, such that $x_2>0$ and for every $n\ge 2$ holds 

$$x_{n+1}=-1+\sqrt[n]{1+nx_n}.$$


Prove consecutively that  $$1)\ \lim_{n\to\infty}x_n=0,\  \ \ 2)\ \lim_{n\to\infty}nx_n=0,\  \ \ 3)\ \lim_{n\to\infty}n^2x_n=4.$$


$\textbf{Proof.}$  $\textbf{1)}$ Clearly all the elements of the sequence are positive. The inequality $-1+\sqrt[n]{1+nx_n}<x_n$ is equivalent to $(1+nx_n)<(1+x_n)^n$, which is seen to be true after expanding, since all the summands on the right are positive. This shows that the sequence is strictly decreasing. Hence

$$0<x_{n+1}=-1+\sqrt[n]{1+nx_n}\le -1+\sqrt[n]{1+nx_2},$$

and since the right hand side clearly tends to $0$ we obtain $\displaystyle\lim_{n\to\infty}x_n=0.$

$\textbf{2)}$ Now $1/x_n\to +\infty$  increasingly and we can use Stolz theorem as follows:

$$\lim_{n\to\infty}nx_n =\lim_{n\to\infty}\frac{n}{\frac{1}{x_n}}=\frac{(n+1)-n}{\frac{1}{x_{n+1}}-\frac{1}{x_n}}=\lim_{n\to\infty}\frac{x_{n+1}x_n}{x_n-x_{n+1}}$$

Now consider the defining equation. It can be rewritten as $(1+x_{n+1})^n=1+n x_n$ hence $ x_n=x_{n+1}+S/n$ where $\displaystyle S=\sum_{k=2}^n{n\choose k}x_{n+1}^k.$

Thus 

$$\lim_{n\to\infty}\frac{x_{n+1}x_n}{x_n-x_{n+1}}=\lim_{n\to\infty}\frac{x_{n+1}\left(x_{n+1}+\frac{S}{n}\right)}{\frac{S}{n}}=\lim_{n\to\infty}\frac{nx_{n+1}^2}{S},$$

where we have used that $x_{n+1}\to 0$. Now this can be rewritten as follows

$$\lim_{n\to\infty}\frac{nx_{n+1}^2}{S}=\lim_{n\to\infty}\frac{n}{{n\choose 2}+S'},$$

 where $\displaystyle S'=\sum_{k=3}^n{n\choose k}x_{n+1}^{k-2}>0.$ This shows that the last limit is $0$.

$\textbf{3)}$  Apply the Stolz theorem in the very same way as above -


$$\lim_{n\to\infty}n^2x_n =\lim_{n\to\infty}\frac{n^2}{\frac{1}{x_n}}=\frac{(n+1)^2-n^2}{\frac{1}{x_{n+1}}-\frac{1}{x_n}}=\lim_{n\to\infty}(2n+1)\frac{x_{n+1}x_n}{x_n-x_{n+1}}=2\lim_{n\to\infty}\frac{n^2x_{n+1}^2}{S}.$$

For convenience we work with the reciprocal limit 

$$\lim_{n\to\infty}\frac{S}{n^2x_{n+1}^2}=\lim_{n\to\infty}\frac{\sum_{k=2}^n{n\choose k}x_{n+1}^k}{n^2x_{n+1}^2}=\frac{1}{2}+\lim_{n\to\infty}\sum_{k=3}^n{n\choose k}\frac{1}{n^2}x_{n+1}^{k-2}.$$

 We would be done if we prove, that the last limit is $0$.  Denote $A_n=\displaystyle \sum_{k=3}^n{n\choose k}\frac{1}{n^2}x_{n+1}^{k-2}$. Clearly $A_n>0$. Fix $\varepsilon>0$. According to $2)$, for large enough $n$ holds $\displaystyle x_{n+1}<\frac{\varepsilon}{n}.$

Thus (after completing to the binomial formula) $$A_n<\sum_{k=3}^n{n\choose k}\frac{\varepsilon^{k-2}}{n^k}=\frac{-n \varepsilon ^2+2 n \left(\frac{n+\varepsilon }{n}\right)^n-2 n \varepsilon -2 n+\varepsilon ^2}{2 n \varepsilon ^2},$$

hence $$\limsup_{n\to\infty}A_n\le \lim_{n\to\infty} \frac{-n \varepsilon ^2+2 n \left(\frac{n+\varepsilon }{n}\right)^n-2 n \varepsilon -2 n+\varepsilon ^2}{2 n \varepsilon ^2}=\frac{-\varepsilon ^2-2 \varepsilon +2 e^{\varepsilon }-2}{2 \varepsilon ^2}.$$

This is true for arbitrary $\varepsilon>0$. Letting $\varepsilon\to 0$ in the last bound we obtain (using Taylor expansion of the exponent near $0$) that  

$$\lim_{\varepsilon\to 0}\frac{-\varepsilon ^2-2 \varepsilon +2 e^{\varepsilon }-2}{2 \varepsilon ^2}=0,$$

whence  $\displaystyle\limsup_{n\to\infty}A_n=0,$ which finishes the proof.

Comments

Post a Comment

Popular posts from this blog

Basel problem type sum, proposed by prof. Skordev

Evaluate $$\sum_{n=1}^{\infty}\frac{1}{2^n n^2}.$$ Proof. The function $$S(x):=\sum_{n=1}^{\infty}\frac{x^n}{ n^2}$$ is well defined for $x\in[-1,1]$ and thus we need to find $S(1/2)$. Using differentiation, we observe that $$S(x)=-\int_0^x\frac{\log(1-t)}{t}\text{d}t.$$ If we make change of variables in the latter integral $t\to 1-t$ we obtain $$S(x)=-\int_{1-x}^1\frac{\log t}{1-t}\text{d}t. \ \ \ \ \ (1)$$ On the other hand, using integration by parts, we obtain $$S(x)=-\int_0^x\log(1-t)\text{d}\log t=-\log t\log(1-t)\Big|_{t=0}^x-\int_0^x\frac{\log t}{1-t}\text{d}t.$$ Notice that $\displaystyle \lim_{t\to 0}\log t \log(1-t)=0$ ($\log (1-t)\sim t$ and then L'Hopital), hence $$S(x)=-\int_0^x\log(1-t)\text{d}\log t=-\log x\log(1-x)-\int_0^x\frac{\log t}{1-t}\text{d}t. \ \ \ \ \ (2)$$ Plugging $x=1/2$ and equating $(1)$ to $(2)$ we obtain $$\int_{1/2}^1\frac{\log(1-t)}{t}\text{d} t-\int_{0}^{1/2}\frac{\log(1-t)}{t}\text{d} t= \log\left(\frac{1}{2}\right)^2.$$ On the ot...
Read more

A problem proposed by prof. Babev

Consider the sequence $$a_n=\int_{n}^{n+1}\frac{\sin^2(\pi t)}{t}\text{d}t.$$ It could be proven that $$\lim_{n\to \infty}n a_n=\frac{1}{2}$$  Now we are going to prove that $$\lim_{n\to\infty}n\left(na_n-\frac{1}{2}\right)=-\frac{1}{4} $$ and show how to derive all such limits. $\textbf{Proof.}$ Since $\displaystyle\int_{n}^{n+1}\sin^2(\pi t)\text{d}{t}=\frac{1}{2}$ (for any $n\in \mathbb{N}$) we can rewrite the limit in question as $$\lim_{n\to \infty}n\left(n\left(\int_{n}^{n+1}\frac{\sin^2(\pi t)}{t}\text{d}t-\int_{n}^{n+1}\frac{\sin^2(\pi t)}{n}\text{d}t\right)\right)=\lim_{n\to \infty}n^2\left(\int_{n}^{n+1}\sin^2(\pi t)\left(\frac{1}{t}-\frac{1}{n}\right)\text{d}t\right)$$ $$=\lim_{n\to \infty}n\left(\int_{n}^{n+1}\sin^2(\pi t)\frac{n-t}{t}\text{d}t\right)$$ Changing the variables $t\to n+s$ and using periodicity of sine, the last simplifies to $$ \lim_{n\to \infty}n\int_{0}^{1}\sin^2(\pi s)\frac{-s}{n+s}\text{d}s$$ Since for $s\in[0,1]$, $\displaystyle \...
Read more