# Thread: Analysis: show xn is a null sequence

1. ## Analysis: show xn is a null sequence

Prove that $\displaystyle x_{n+1}-1/2x_n\to 0 \Rightarrow x_n\to 0$.

My approach:
Let $\displaystyle \varepsilon>0$. Then there exists an $\displaystyle N\in \Bbb N$ s.t. $\displaystyle \forall n\geq N, |x_{n+1}-1/2x_n|<\varepsilon$. It follows that $\displaystyle -\varepsilon\leq -\varepsilon+1/2x_n<x_{n+1}<\varepsilon \leq \varepsilon+1/2x_n$. Then for any $\displaystyle n\geq N$, we have $\displaystyle |x_{n+1}|<\varepsilon$.
But I have assumed that $\displaystyle \forall n\in \Bbb N, x_n\geq 0$ which may not be the case. Can I argue that when $\displaystyle x_n<0$, we have $\displaystyle -\varepsilon+1/2x_n<-\varepsilon<x_{n+1}<\varepsilon+1/2x_n<\varepsilon$? Is my proof correct?

Thank you!

2. ## Re: Analysis: show xn is a null sequence

Hey ach4124.

You have "sandwiched" your epsilon so that the absolute value of your x_n is less than it (which is the thing about epsilons often going to zero even if they don't actually approach it "technically") which is important.

The argument of the epsilon has to be correct in terms of xn satisfying that inequality so from that point of view it looks good.

You could also use things like the triangle inequality but I'm not sure what your teacher has instructed (in analysis, you can use lots of existing identities to prove these sorts of things).

3. ## Re: Analysis: show xn is a null sequence

Are you actually required to use $\displaystyle \delta-\epsilon$? I would write that $\displaystyle x_{n+1}=\tfrac12x_n$ and use that to write a closed form expression for $\displaystyle x_n$ in terms of $\displaystyle n$.

4. ## Re: Analysis: show xn is a null sequence

Thank you for your response. At second glance, I think my proof is flawed! In my proof, $\displaystyle 1/2x_n$ does not 'hold any place', so I guess we can extend the statement to $\displaystyle x_{n+1}-bx_n\to 0\Rightarrow x_n\to 0$ for $\displaystyle b\in \Bbb Q$, but when $\displaystyle b=1$, $\displaystyle x_n=\sum 1/n$ shows that the statement does not hold.

I couldn't prove it using triangle inequality $\displaystyle |x_{n+1}-1/2x_n|\leq |x_{n+1}|+1/2|x_n|$. Yet, $\displaystyle |x_{n+1}-1/2x_n|<\varepsilon$ does not imply $\displaystyle |x_{n+1}|+1/2|x_n|<\varepsilon$?

I should use the reverse one. $\displaystyle ||x_{n+1}|-1/2|x_n||\leq |x_{n+1}-1/2x_n|<\varepsilon$, so $\displaystyle -\varepsilon\leq -\varepsilon+1/2|x_n|<|x_{n+1}|<-\varepsilon\leq \varepsilon+1/2|x_n|$. This is a bit nicer than my proof because we don't need to care about the sign of $\displaystyle x_n$.

5. ## Re: Analysis: show xn is a null sequence

Thanks, Archie. That's a nice method!

6. ## Re: Analysis: show xn is a null sequence

As an alternative:
$\displaystyle x_{n+1}-\tfrac12x_n=0 \implies x_{n+1}-x_n=-\tfrac12x_n$, so the sequence $\displaystyle \{x_n\}$ is decreasing when $\displaystyle x_n$ is positive and increasing when $\displaystyle x_n$ is negative.

Also $\displaystyle x_{n+1}=\tfrac12x_n$ tells us that $\displaystyle x_{n+1}$ has the same sign as $\displaystyle x_n$.

Thus, the sequence $\displaystyle \{x_n\}$ is either increasing and bounded above (by zero), or it is decreasing and bounded below (by zero), or (trivially) every $\displaystyle x_n$ is equal to zero. This tells us that the series converges.

When a series converges, $\displaystyle \lim_{n \to \infty} x_{n+1}-x_{n} =0$. But since $\displaystyle x_{n+1}=\tfrac12x_n$, we have $\displaystyle \lim_{n \to \infty} \tfrac12x_n-x_{n} =\lim_{n \to \infty} -\tfrac12x_n =0$ and so $\displaystyle \lim_{n \to \infty} x_{n} =0$ as required.

7. ## Re: Analysis: show xn is a null sequence

Was the original question supposed to be $\displaystyle x_{n+1}-\tfrac12x_n=0$ as I read to be, or $\displaystyle x_{n+1}-\tfrac12x_n \to 0$ as you wrote?

8. ## Re: Analysis: show xn is a null sequence

It is $\displaystyle x_{n+1}-\frac{1}{2}x_n\to 0$.

9. ## Re: Analysis: show xn is a null sequence

That's unusual. You might want to check that with your teacher. I'll have to think about what can be done with it though.

10. ## Re: Analysis: show xn is a null sequence

I found this exercise in Selected Problems in Real Analysis, an AMS book. The solution given involves sums: "Let $\displaystyle \tilde{x}_n=x_n-x_{n-1}/2$. Then $\displaystyle x_n-\frac{x_1}{2^{n-1}}=\sum_{1<k\leq n} \frac{\tilde{x}_k}{2^{n-k}}=\sum_{1<k\leq m}+\sum_{m<k\leq n}$. Since $\displaystyle \tilde{x}_k\to 0$, for large m the second sum is small for all $\displaystyle n\geq m$. For fixed m the first sum is arbitrary small if n is sufficiently large."
How does that prove $\displaystyle x_n\to 0$? Does my argument using reverse triangle inequality suffice to prove the statement? I don't understand why we need to consider sums.