# I have a problem

1. Jan 8, 2005

### Final

$$x_{n}+px_{n-1}$$ converges $$\Leftrightarrow x_{n}$$ converges.
For which p it is true?

Thanks
Final

2. Jan 9, 2005

### Rogerio

"Xn converges -> Xn + pXn-1 converges" is always true.
"Xn + pXn-1 converges -> Xn converges" is true if |p| < 1

3. Jan 9, 2005

### learningphysics

I'm guessing there are two series. The first being $$x_{n}$$ and the second being defined in terms of the first as: $$y_{n}=x_{n}+px_{n-1}$$

Assume $$x_{n}$$ converges. So there exists a finite L such that:
$$\lim_{n \rightarrow \infty} x_{n} = L$$

To see whether or not $$y_{n}$$ converges, calculate the limit:
$$\lim_{n \rightarrow \infty} x_{n} + px_{n-1}$$

$$= \lim_{n \rightarrow \infty} x_{n} + p \lim_{n \rightarrow \infty} x_{n-1}$$

$$= L + pL$$

L+pL exists and it is finite for all p, so $$y_{n}$$ converges.

So we've shown
$$x_{n}$$ converges $$\rightarrow x_{n} + px_{n-1}$$ converges for all p.

Assume $$y_{n}$$ converges. So there is a finite M such that:
$$\lim_{n \rightarrow \infty} x_{n} + px_{n-1} = M$$

we can rewrite the left side:
$$\lim_{n \rightarrow \infty} x_{n} + p \lim_{n \rightarrow \infty} x_{n-1} = M$$

$$\lim_{n \rightarrow \infty} x_{n} + p \lim_{n \rightarrow \infty} x_{n} = M$$

$$(1+p) \lim_{n \rightarrow \infty} x_{n} = M$$

$$\lim_{n \rightarrow \infty} x_{n} = M/(1+p)$$

M/(1+p) exists and is finite for all p not equal to -1.

$$x_{n} + px_{n-1}$$ converges $$\rightarrow x_{n}$$ converges for all p not equal -1

Last edited: Jan 9, 2005
4. Jan 9, 2005

### Muzza

That's only true if x_n converges, which is what you were trying to prove... ;)

5. Jan 9, 2005

### mathman

Example where yn series converges and xn does not is as follows: xn =1 for even n and -1 for odd n, for p =1, all yn =0, while the xn series oscillates between 1 and 0 (assuming start with n=0).

6. Jan 9, 2005

### learningphysics

Yes, you're right. All I did was show that if the limit exists then it is M/(p+1) and that the limit does not exist for p=-1. But I did not show where the limit does exist.

Anyway, please ignore my solution... Only look if you want to know how NOT to do calculus!

Last edited: Jan 9, 2005
7. Jan 10, 2005

### mathman

It is easy to construct an example for any p, where |p|<=1.
Let xn=(-1/p)n. Then the x series diverges or oscillates, while the y series is always 0.