Convergence of a sequence depending on a recurrence relation

Question

I would require some help with a series problem my math teacher gave me. I am first year of university. The problem is as follows:

Let there be $x_1 \gt 0$ and the following recurrent relation :

$$x_{n+1} = x_n + \frac{n^2}{x_1+x_2+...+x_n}$$

Study the nature of convergence for $\frac{x_n}{n}$.

Thanks in advance!

Answer 1

The sequence $\{x_n\}_{n\geq 1}$ is clearly positive and increasing. It follows that

$$ x_{n+1}=x_n+\frac{n^2}{x_1+\ldots+x_n}\geq x_n+\frac{n}{x_n} \tag{A}$$ $$ x_{n+1}^2 \geq x_n^2 + 2n,\qquad x_n\geq (n-1) \tag{B} $$ $$ x_{n+1}\leq x_n +\frac{n^2}{0+1+\ldots+(n-1)},\qquad x_n\leq x_0+2n\tag{C} $$ so $\frac{x_n}{n}$ is bounded. Assuming $\left\{\frac{x_n}{n}\right\}_{n\geq 1}$ is convergent, by the Cesàro-Stolz theorem we have

$$ L=\lim_{n\to +\infty}\frac{x_n}{n}=\lim_{n\to +\infty}\frac{n^2}{x_1+\ldots+x_n}=\lim_{n\to +\infty}\frac{2n+1}{x_{n+1}}=\frac{2}{L}\tag{D}$$ hence $L=\color{red}{\sqrt{2}}$.

Now, let us prove that the limit exists. By $(B)$ and $(C)$ $x_n$ is bounded between $n+O(1)$ and $2n+O(1)$, so $a_n = \frac{x_n}{n}$ is bounded between $1-\varepsilon$ and $2+\varepsilon$. We have

$$ n(a_{n+1}-a_n) = \frac{2n}{(n+1)}\cdot\frac{n(n+1)/2}{a_1+ 2a_2+\ldots + na_n}-a_{n+1} \tag{E}$$ where $\frac{n(n+1)/2}{a_1+ 2a_2+\ldots + na_n}$ is the reciprocal of a weighted mean of $a_1,\ldots,a_n$. It follows that $\{a_n\}_{n\geq 1}$ is "almost constant", in the sense that $a_{n+1}-a_n = O\left(\frac{1}{n}\right)$. Let us assume that $\{a_n\}_{n\geq 1}$ has a subsequence $\{a_{n_k}\}_{k\geq 1}$ convergent to $c\neq \sqrt{2}$. By $(E)$ $$ a_{n_k+1}-a_{n_k} = \frac{1}{n_k}\left(\frac{2}{c}-c+o(1)\right)\tag{F} $$ so the weighted mean of $a_1,a_2,\ldots,a_{n_k},a_{n_k+1}$ is also close to $c$ and $(F)$ continues to hold by replacing $n_k$ with $n_k+1$. On the other hand, by summing multiple instances of $(F)$ we violate the boundedness of $\{a_n\}_{n\geq 1}$, since the harmonic series is divergent. It follows that $\sqrt{2}$ is the only limit point of $\{a_n\}_{n\geq 1}$.