The radius of convergence of a series

In summary: We look at the series ##\sum a_n \, t^n##. Apparently there is a theorem that the radius of convergence is equal to ##1/\lim {a_{n+1}/a_n}##.
  • #1
310
54
Homework Statement
look at the image
Relevant Equations
r=an+1/an
Greetings!
I have a problem with the solution of that exercice
1642767460771.png

1642767491965.png


I don´t agree with it because if i choose to factorise with 6^n instead of 2^n will get 5/6 instead thank you!
 
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  • #2
Amaelle said:
if i choose to factorise with 6^n instead of 2^n will get 5/6
Can you show us how you do that ?

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  • #3
thank you
here it is
20220121_140409756.jpg
 
  • #4
So your ##t = x+{16\over 6}##. What is your ##a_n## ? It looks as if the ##2^n## in the denominator has disappeared ...

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  • #5
BvU said:
So your ##t = x+{16\over 6}##. What is your ##a_n## ? It looks as if the ##2^n## in the denominator has disappeared ...

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of course according to the asymptotic aprroximation 5^n >2^n so I got rid of it
 
  • #6
So if you factorize with, say, 24, you get ##\displaystyle{5\over 24}## ?

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  • #7
thank you!
First could you confirm that my assymptotic approximation is correct?
second my original approch was to put y=6x+16 which gives me a raduis of 5/6 I just don´t understand this book approach
 
  • #8
##2^n(3x + 8)^n = 6^n(x + \frac 8 3)^n##. Of course ##x + 8/3## is the same as ##x + 16/6##, but why leave it in this unsimplified form?

In your work using the Ratio Test, you have omitted the ## 6^n(x + \frac 8 3)^n## part, which is a mistake.
 
  • #9
thank you but
the ration test is an+1/an and (x+8/3)^n is not part of it's part of t^n
 
  • #10
Amaelle said:
the ration test is an+1/an

The Ratio Test for a power series definitely includes the part with the x + 8/3 factor. It's how you determine the radius of convergence.
For example, for the power series ##\displaystyle \sum_{i = 0}^\infty a_n(x - c)^n##, the Ratio Test would be to evaluate this limit:
##\displaystyle \lim_{n \to \infty} \frac {a_{n+1}(x - c)^{n+1}}{a_n(x - c)^n}##.

BTW, when you write "an+1/an" this would be interpreted by most as ##a_n + \frac 1 {a_n}##. If you don't use LaTeX, write this expression as (an + 1)/(an).
 
  • #11
Amaelle said:
thank you but
the ration test is an+1/an and (x+8/3)^n is not part of it's part of t^n
That is why I asked you to post your ##a_n##. What I see in #3 is wrong.

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  • #12
what´s wrong with it? I posted all the elements I have and done, I would be grateful if you could point out the problem clearely.
 
  • #13
Amaelle said:
what´s wrong with it? I posted all the elements I have and done, I would be grateful if you could point out the problem clearely.
Excuse me ? Your first ##=## sign is wrong, the second ##=## sign is wrong. And you can not just 'get rid of' the 2^n.

BvU said:
What is your ##a_n## ?

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  • #14
BvU said:
And you can not just 'get rid of' the 2^n.
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is not 5^n+2^n ~ 5^n assymptotically?
 
  • #15
Amaelle said:
what´s wrong with it? I posted all the elements I have and done, I would be grateful if you could point out the problem clearely.
The general term of you series is this:
$$\frac{n \log^{2021}(n)}{2^n + 5 ^n}6^n(x + 16/6)^n$$
In the second line of your work in post #3 you have this, I think -- your work is hard to read in places:
$$\frac{ (n + 1) \log^{2022}(n) 6^{n+1} 5^n} {5^{n + 1} n \log^{2021}(n) 6^n } $$
This is wrong on several counts.
1. The log function doesn't get bumped up another power. It should be ##\log^{2021}(n + 1)##.
2. As BvU said, you can't just throw away the ##2^n## term. Instead, factor it out to get ##2^n(1 + (5/2)^n)##.
3. You should also not throw away the x + 8/3 factor.

BvU said:
And you can not just 'get rid of' the 2^n.
Yes
 
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  • #16
Amaelle said:
is not 5^n+2^n ~ 5^n assymptotically?
No. The difference gets bigger (by a factor of 2) with every step :smile:

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  • #17
Mark44 said:
3. You should also not throw away the x + 8/3 factor.
We look at the series ##\sum a_n \, t^n##. Apparently there is a theorem that the radius of convergence is equal to ##1/\lim {a_{n+1}/a_n}##.

##\ ##
 
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  • #18
BvU said:
We look at the series ##\sum a_n \, t^n##. Apparently there is a theorem that the radius of convergence is equal to ##1/\lim {a_{n+1}/a_n}##.

##\ ##
yes
 
  • #19
BvU said:
We look at the series ##\sum a_n \, t^n##. Apparently there is a theorem that the radius of convergence is equal to ##1/\lim {a_{n+1}/a_n}##.
For convergence, the ratio test requires
$$\lim_{n \to \infty} \left\lvert \frac{a_{n+1}t^{n+1}}{a_n t^n} \right\rvert = \lvert t \rvert \lim_{n \to \infty} \left\lvert \frac{a_{n+1}}{a_n} \right\rvert < 1,$$ so we have
$$\lvert t \rvert < \lim_{n \to \infty} \left\lvert \frac{a_{n}}{a_{n+1}} \right\rvert.$$

In the solution, the value of the limit is claimed to be the radius of convergence, but presumably, when we talk about the radius of convergence, we're talking about the size of the interval in terms of ##x##, not ##t##. It seems to me you still need to rescale the value of the limit depending on how ##t## is defined in terms of ##x##.
 
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  • #20
vela said:
For convergence, the ratio test requires
$$\lim_{n \to \infty} \left\lvert \frac{a_{n+1}t^{n+1}}{a_n t^n} \right\rvert = \lvert t \rvert \lim_{n \to \infty} \left\lvert \frac{a_{n+1}}{a_n} \right\rvert < 1,$$ so we have
$$\lvert t \rvert < \lim_{n \to \infty} \left\lvert \frac{a_{n}}{a_{n+1}} \right\rvert.$$

In the solution, the value of the limit is claimed to be the radius of convergence, but presumably, when we talk about the radius of convergence, we're talking about the size of the interval in terms of ##x##, not ##t##. It seems to me you still need to rescale the value of the limit depending on how ##t## is defined in terms of ##x##.

Thanks a million vela,
that was my question and my concern!
indeed after rescaling it we find 5/6.
 
  • #22
Amaelle said:
Thanks a million vela,
that was my question and my concern!
indeed after rescaling it we find 5/6.
Same exercise with ##\displaystyle 1\over 4^n + 5^n## instead of ##\displaystyle 1\over 2^n + 5^n\ ##: do you get the same answer or a different one ?

@vela: yes, the book answer leaves unmentioned that the ##\rho## found applies to ##3x+8##.

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  • #23
BvU said:
book answer leaves unmentioned that the ρ found applies to 3x+8.
Which is why I kept saying that we need the 3x + 8 term (or x + 8/3) to find the interval of convergence.
 
  • #24
BvU said:
Same exercise with ##\displaystyle 1\over 4^n + 5^n## instead of ##\displaystyle 1\over 2^n + 5^n\ ##: do you get the same answer or a different one ?

@vela: yes, the book answer leaves unmentioned that the ##\rho## found applies to ##3x+8##.

##\ ##
Yes I got the same result (without doing the problematic assymptotic approximation you pointed out)
 

What is the radius of convergence of a series?

The radius of convergence of a series is a value that determines the interval of values for which the series will converge. It is denoted by R and is calculated using the ratio test or the root test.

How is the radius of convergence of a series calculated?

The radius of convergence, R, is calculated by taking the limit of the absolute value of the ratio of consecutive terms in the series. If this limit is less than 1, then the series will converge. The radius of convergence can also be calculated using the root test, where the limit of the nth root of the absolute value of the nth term is taken.

What does the radius of convergence tell us about a series?

The radius of convergence tells us the interval of values for which the series will converge. If a value falls within this interval, then the series will converge. If a value falls outside of this interval, then the series will diverge.

Can the radius of convergence of a series be negative?

No, the radius of convergence must be a positive value. This is because it represents the distance from the center of the interval of convergence to either end. A negative radius would not make sense in this context.

What happens if the radius of convergence is infinite?

If the radius of convergence is infinite, then the series will converge for all values of the variable. This means that the interval of convergence is the entire real number line. This is usually the case for simple series like geometric series or power series with only one variable.

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