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Radius of Convergence

I need to show that the radius of convergence of anxnanx^n is the same as the radius of convergence of nanxn−1na_nx^{n-1}

Would it be ok to use the ratio test as below (the limit is as n tends to infinity, not sure of the latex for limits)?

lim((n+1)an+1xn−1nanxn\frac{(n+1) a_{n+1} x^{n-1}}{na_nx^n})=x*lim(an+1an\frac{a_{n+1}}{a_n})

which is the same as for anxna_nx^n
Reply 1
Avatar for Mark13
Mark13
Original post by james22
I need to show that the radius of convergence of anxnanx^n is the same as the radius of convergence of nanxn−1na_nx^{n-1}

Would it be ok to use the ratio test as below (the limit is as n tends to infinity, not sure of the latex for limits)?

lim((n+1)an+1xn−1nanxn\frac{(n+1) a_{n+1} x^{n-1}}{na_nx^n})=x*lim(an+1an\frac{a_{n+1}}{a_n})

which is the same as for anxna_nx^n


Yep that looks good.
Reply 2
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james22
OP
16
Original post by Mark13
Yep that looks good.


OK thanks, I'm just unsure as all the proofs I've found online are much more complicated than that.
Reply 3
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Noble.
17
I don't know why, but I have the feeling there's an issue with showing the derivative of a power series has the same radius of convergence just simply using the ratio test like this. This is proven in our lecture notes though (Analysis II Page 50) where they show that R′≤RR' \leq R and R′≥RR' \geq R
Reply 4
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james22
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16
Original post by Noble.
I don't know why, but I have the feeling there's an issue with showing the derivative of a power series has the same radius of convergence just simply using the ratio test like this. This is proven in our lecture notes though (Analysis II Page 50) where they show that R′≤RR' \leq R and R′≥RR' \geq R


I just found the proof in my notes, and it is very similar to the one here, the ratio test is much simpler which is why I am now unsure.
It's definitely not good! What happens if the limit of the ratios is one?
Indeed, what happens if the ratios don't even converge?
(edited 10 years ago)
Reply 6
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DFranklin
18
Original post by IrrationalNumber
It's definitely not good! What happens if the limit of the ratios is one?
I don't see this is a problem.

Indeed, what happens if the ratios don't even converge?
This is, however.

You can avoid this problem using http://en.wikipedia.org/wiki/Root_test
Reply 7
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james22
OP
16
Original post by IrrationalNumber
It's definitely not good! What happens if the limit of the ratios is one?
Indeed, what happens if the ratios don't even converge?



Original post by DFranklin
I don't see this is a problem.

This is, however.

You can avoid this problem using http://en.wikipedia.org/wiki/Root_test


That makes sense, having only done questions where the ratio test always converges to something I didn't even consider that it only worked when the sequence actually converged.
Reply 8
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DFranklin
18
Original post by james22
That makes sense, having only done questions where the ratio test always converges to something I didn't even consider that it only worked when the sequence actually converged.
It's not actually that uncommon for the ratio not to converge (any series with only odd or even powers won't have the ratio converge), although it's usually "obvious" how to fix it if you use a moment's common sense.

But to be honest, I was also in the "yeah, that's fine" mindset until I saw Irrational's post.
Original post by DFranklin
I don't see this is a problem.

Yep, you're right. The convergence issue remains though.

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