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.