# Checking the convergence of system of sequences.

Let $\{u_k\}_{k\in \mathbb{N}}$ and $\{v_k\}_{k\in \mathbb{N}}$ be the two sequences given by $$\begin{cases} u_0=v_0=1,&\\ u_{k+1}=\left(1-\frac{1}{(k+1)^2}\right)u_k+\left(\frac{1}{k+1}\right)v_k, & (k\in \mathbb{N})\\ v_{k+1}=\left(-\frac{1}{k+1}\right)u_k+\left(1-\frac{1}{(k+1)^2}\right)v_k. & (k\in \mathbb{N}) \end{cases}$$ Checking the convergence of the sequences $\{u_k\}_{k\in \mathbb{N}}$ and $\{v_k\}_{k\in \mathbb{N}}$.

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What have you tried? –  Alex Becker Mar 19 '12 at 22:29
Dear Alex Becker. I tried to do it. I guess that both the sequences are not convergent. Thank you for your consideration of my question. –  impartialmale Mar 19 '12 at 22:45

Let $z_k = u_k + i v_k$. The recursion says $z_{k+1} = a_{k+1} z_k$, where $a_k = 1 - 1/k^2 - i/k$. Now $a_k = r_k e^{i\theta_k}$ where $r_k = |a_k| = \sqrt{1 - 1/k^2 + 1/k^4}$ and $\theta_k = -\arctan\left(\frac{1/k}{1 - 1/k^2}\right)$. Thus $z_n = z_0 \prod_{k=1}^n a_k = \left(\prod_{k=1}^n r_k \right) \exp\left(i \sum_{k=1}^n \theta_k\right)$.
We have $r_k = 1 - 1/(2 k^2) + O(1/k^4)$ and $\theta_k = -1/k + O(1/k^3)$ as $k \to \infty$. The infinite product $\prod_{k=1}^\infty r_k$ converges to a nonzero limit $R$, so $|z_n| \to R |z_0|$ as $n \to \infty$, but $\sum_{k=1}^\infty \theta_k$ diverges, so $\arg(z_n)$ diverges, and so do the sequences $u_n$ and $v_n$. In the limit as $n \to \infty$, the points $z_n$ spiral infinitely many times around the circle of radius $R |z_0|$ centred at $0$.
Thank you Robert Israel for your nice solution. I want to ask you more. If we take a point $\bar{z}$ in the circle of radius $R|z_0|$ centered at $0$, could we choose a subsequence $\{z_{n_i}\}\subset \{z_n\}$ converging to $\bar{z}$. –  impartialmale Mar 20 '12 at 5:08
I am still confused. Why $\arg(z_n)$ diverges implies both $\{u_n\}$ and $\{v_n\}$ diverge? Let $z_n=u_n+iv_n=r_n(\cos(\theta_n)+i\sin(\theta_n))=r_ne^{i\theta_n}$. Let $r_n=r$ (constant) and $\theta_n=2n\pi$ for all $n\in \mathbb{N}$. Then $\theta_n$ is not convergent and $u_n=r$, $v_n=0$ for all $n\in\mathbb{N}$. This means that both $\{u_n\}$ and $\{v_n\}$ are convergent. –  impartialmale Mar 20 '12 at 10:01
Maybe I should have said $\text{Arg}(z_n)$ (i.e. the version of arg in the interval $(-\pi,\pi]$) diverges. –  Robert Israel Mar 20 '12 at 18:43