# Thread: Analysis: show xn is a null sequence

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

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

My approach:
Let $\varepsilon>0$. Then there exists an $N\in \Bbb N$ s.t. $\forall n\geq N, |x_{n+1}-1/2x_n|<\varepsilon$. It follows that $-\varepsilon\leq -\varepsilon+1/2x_n. Then for any $n\geq N$, we have $|x_{n+1}|<\varepsilon$.
But I have assumed that $\forall n\in \Bbb N, x_n\geq 0$ which may not be the case. Can I argue that when $x_n<0$, we have $-\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 $\delta-\epsilon$? I would write that $x_{n+1}=\tfrac12x_n$ and use that to write a closed form expression for $x_n$ in terms of $n$.

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, $1/2x_n$ does not 'hold any place', so I guess we can extend the statement to $x_{n+1}-bx_n\to 0\Rightarrow x_n\to 0$ for $b\in \Bbb Q$, but when $b=1$, $x_n=\sum 1/n$ shows that the statement does not hold.

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

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

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

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

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

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

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

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

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

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

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 $\tilde{x}_n=x_n-x_{n-1}/2$. Then $x_n-\frac{x_1}{2^{n-1}}=\sum_{1. Since $\tilde{x}_k\to 0$, for large m the second sum is small for all $n\geq m$. For fixed m the first sum is arbitrary small if n is sufficiently large."
How does that prove $x_n\to 0$? Does my argument using reverse triangle inequality suffice to prove the statement? I don't understand why we need to consider sums.