# Math Help - Inductive sequences

1. ## Inductive sequences[Unsolved]

Inductively define a sequence $\{x_n\}$ by $x_1 = 1$ and $x_n = x_{n-1} - \frac{(-1)^n}{n}$. Show that $\{x_n\}$ converges.

I'm having trouble proving this, my idea was to show that this sequence was Cauchy and hence convergent(since we're in $\mathbb{R}$). However I'm having difficulty showing its Cauchy as for arbitrary $\epsilon > 0$, I need $|x_m - x_n| < \epsilon$ where this holds for some integer $N$, and $\forall m, n \geq N$. My problem is with how to choose an $N$ such that $|x_{m-1} - \frac{(-1)^m}{m} - x_{n-1} + \frac{(-1)^n}{n}| < \epsilon$ holds for arbitrary $\epsilon > 0$.

Also another similar question:
Inductively define a sequence $\{s_n\}$ by $s_1 = 3$ and $s_{n+1} = \frac{2}{3}s_n + \frac{4}{3s_n}$. Prove $\{s_n\}$ converges and evaluate the limit.

My idea to show that it converges was to show that its monotonically decreasing and bounded. And then I know the limit is 2, since I played around with it on my calculator. However I had problems when I tried to use induction to show that its monotonically decreasing, the equations get really messy and I never get anything nice I can use.

Also a similar things happens when, for abitrary $\epsilon > 0$ I try find $N$ such that $\forall n \geq N$, $|s_n - 2| < \epsilon$. The definition of $\{s_n\}$ is not in terms of $n$, so I'm not sure how to find an $N$ that meets the requirement.

Any hints would be greatly appreciated.

2. Originally Posted by utopiaNow
Inductively define a sequence $\{x_n\}$ by $x_1 = 1$ and $x_n = x_{n-1} - \frac{(-1)}{n}$. Show that $\{x_n\}$ converges.

I'm having trouble proving this, my idea was to show that this sequence was Cauchy and hence convergent(since we're in $\mathbb{R}$). However I'm having difficulty showing its Cauchy as for arbitrary $\epsilon > 0$, I need $|x_m - x_n| < \epsilon$ where this holds for some integer $N$, and $\forall m, n \geq N$. My problem is with how to choose an $N$ such that $|x_{m-1} - \frac{(-1)}{m} - x_{n-1} + \frac{(-1)}{n}| < \epsilon$ holds for arbitrary $\epsilon > 0$.
Do you mean $x_{n-1}-\frac{(-1)^{\color{red}n}}{n}$?

If so, look what happens when you write out the first few terms:

$x_1=1$
$x_2=1-\frac{1}{2}$
$x_3=1-\frac{1}{2}+\frac{1}{3}$
...

So $x_n$ will be the $n$th partial sum of $\sum_{k=1}^{\infty}\frac{(-1)^{n+1}}{n}$, which converges by the alternating series test (to $\ln2$, but that's irrelevant for this problem).

3. Originally Posted by redsoxfan325
Do you mean $x_{n-1}-\frac{(-1)^{\color{red}n}}{n}$?
Oops, my mistake. Yes I did mean that. I read over what I had typed so many times and still I couldn't see my typo. Sorry about that. I'll fix it up top.