1. ## Show sequence convergence

1. Let $\displaystyle (x_n)$ be a sequence with $\displaystyle x_1 = 1$ and $\displaystyle x_n = x_{n-1} - \frac{(-1)^n}{n}$. Prove convergence.

I tried to prove this is cauchy, but couldn't make much progress.

2. Let $\displaystyle (s_n)$ be a sequence with $\displaystyle s_1 = 3$, $\displaystyle s_{n+1} = \frac{2}{3}s_n + \frac{4}{3s_n}$. Prove convergence and evaluate the limit.

I know the limit is 2, but I don't understand how to show this using an $\displaystyle \epsilon$-proof.

2. Originally Posted by chrischen88
1. Let $\displaystyle (x_n)$ be a sequence with $\displaystyle x_1 = 1$ and $\displaystyle x_n = x_{n-1} - \frac{(-1)^n}{n}$. Prove convergence.

I tried to prove this is cauchy, but couldn't make much progress.

2. Let $\displaystyle (s_n)$ be a sequence with $\displaystyle s_1 = 3$, $\displaystyle s_{n+1} = \frac{2}{3}s_n + \frac{4}{3s_n}$. Prove convergence and evaluate the limit.

I know the limit is 2, but I don't understand how to show this using an $\displaystyle \epsilon$-proof.
1. Write out the first few terms:

$\displaystyle x_1=1$
$\displaystyle x_2=1-\frac{1}{2}$
$\displaystyle x_3=1-\frac{1}{2}+\frac{1}{3}$
...
$\displaystyle x_n=\sum_{k=1}^n\frac{(-1)^{n+1}}{n}$

The alternating series test says that this converges. (It happens to converge to $\displaystyle \ln2$, but that's neither here nor there.)

2. You don't need an $\displaystyle \epsilon$-proof. I would try to prove that it's monotonically decreasing, and convergence would follow from there (because it's obviously bounded below by $\displaystyle 0$). You should be able to do it by induction, though I won't go into the details, because it's likely to be tedious. Once you know that it converges, finding the number to which it converges is easy:

$\displaystyle \lim_{n\to\infty}s_{n+1}=\lim_{n\to\infty}\left(\f rac{2}{3}s_n+\frac{4}{3s_n}\right)\implies s=\frac{2}{3}s+\frac{4}{3s}\implies\frac{s^2-4}{3s}=0\implies s=2$

According to the equation, $\displaystyle s=-2$ is also a solution, but we can ignore it because $\displaystyle \{s_n\}$ is bounded below by $\displaystyle 0$.