Which p values result in the convergence of the series x_{n}+px_{n-1}?

[Final]

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

Thanks
Final

 

[Rogerio]

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

 

[learningphysics]

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

Thanks
Final

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

 

[Muzza]

...

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

...

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

 

[mathman]

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

 

[learningphysics]

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

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!

 

[mathman]

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