• Poetria
In summary, the conversation discusses the concept of radius of convergence and the given limit involving two power series. The radius of convergence for both series is 1, and when taking the limit, the expression inside should be manipulated before taking the limit. The difference between the two series cancels out except for the part with ##n=0## to ##n=252##, which converges for every value of x. However, in the limit, the difference only makes sense if ##|x|<1##.
Poetria

## Homework Statement

##f(x)=\sum_{n=0}^\infty x^n##
##g(x)=\sum_{n=253}^\infty x^n##

The radius of convergence of both is 1.

## \lim_{N \rightarrow +\infty} \sum_{n=0}^N x^n - \sum_{n=253}^N x^n##

2. The attempt at a solution

I got:
## \frac {x^{253}} {x-1}+\frac 1 {1-x}## for ##|x| \lt 1##

so the radius of convergence of the sum of these power series is just the same and equals 1?

I assume you are asking for the radius of convergence of the given limit, so: No. Manipulate the expression inside the limit before taking the limit.

Poetria
Orodruin said:
I assume you are asking for the radius of convergence of the given limit, so: No. Manipulate the expression inside the limit before taking the limit.

I guess I should shift the index?

##\lim_{N \rightarrow +\infty} \sum_{n=253}^N x^{n-253}-x^n##

Then radius of convergence would be infinite?

That is not correct. Note that the series in your expression have a different number of terms!

Poetria
If I understand this correctly the difference between the two is:

## \sum_{n=0}^{253} x^n##

Perhaps they cancel each other except for this part?

Yeah I have noticed the number of terms isn't equal. :( Hm

Poetria said:
If I understand this correctly the difference between the two is:

## \sum_{n=0}^{253} x^n##

Perhaps they cancel each other except for this part?

They do indeed (except the new sum just goes to ##n = 252##). So does that sum depend on ##N##?

Poetria
Orodruin said:
They do indeed (except the new sum just goes to ##n = 252##). So does that sum depend on ##N##?
Not at all. That's clear. :) Oh yes to 252. Of course.

Therefore it converges for every x, am I right?

Yes.

Poetria
Orodruin said:
Yes.

Great. :) Many thanks. :)

Poetria said:
Therefore it converges for every x, am I right?

Well, it is just a finite polynomial in x, so why would it ever diverge?

Poetria
Ray Vickson said:
Well, it is just a finite polynomial in x, so why would it ever diverge?

:) True. I guess the concept of radius of convergence seems a bit tricky to me and I am overegging the pudding. :( Wolfram Alpha doesn't help.

Poetria said:
:) True. I guess the concept of radius of convergence seems a bit tricky to me and I am overegging the pudding. :( Wolfram Alpha doesn't help.

In a way, this question is trickier than it looks. For any finite ##N > n## the difference ##S_n(x) = \sum_{k=0}^N x^k - \sum_{k=n+1}^N x^k## is just ##1+x+x^2 + \cdots + x^n##, a finite polynomial that exists and makes sense for all ##x##. However, when we go to the limit ##N \to \infty##, the difference ##\sum_{k=0}^{\infty} x^k - \sum_{k=n+1}^{\infty} x^k## only makes sense when each of the sums converge, so only if ##|x| < 1##. On the other hand, we have##1+x+x^2 + \cdots + x^n = \sum_{k=0}^N x^k - \sum_{k=n+1}^N x^k## for ALL ##N > n##, so is true also in the limit! In other words ##\lim (\sum - \sum )## is always OK but ## (\lim \sum) - (\lim \sum) ## might not be.

Last edited:
Poetria
Ray Vickson said:
In a way, this question is trickier than it looks
I would argue that this is the entire point of the question...

Poetria
Ray Vickson said:
In a way, this question is trickier than it looks. For any finite ##N > n## the difference ##S_n(x) = \sum_{k=0}^N x^k - \sum_{k=n+1}^N x^k## is just ##1+x+x^2 + \cdots + x^n##, a finite polynomial that exists and makes sense for all ##x##. However, when we go to the limit ##N \to \infty##, the difference ##\sum_{k=0}^{\infty} x^k - \sum_{k=n+1}^{\infty} x^k## only makes sense when each of the sums converge, so only if ##|x| < 1##. On the other hand, we have##1+x+x^2 + \cdots + x^n = \sum_{k=0}^N x^k - \sum_{k=n+1}^N x^k## for ALL ##N > n##, so is true also in the limit! In other words ##\lim (\sum - \sum )## is always OK but ## (\lim \sum) - (\lim \sum) ## might not be.
Many thanks for great explanation. :) I spent a lot of time thinking about it. :) Now I see what I didn't get.

## 1. What is a power series?

A power series is an infinite series of the form ∑ cn(x-a)n, where cn and a are constants and x is a variable.

## 2. What is the radius of convergence for a power series?

The radius of convergence for a power series is the distance from the center of the series (a) to the nearest point where the series still converges.

## 3. How is the radius of convergence determined?

The radius of convergence can be determined by using the ratio test, which involves taking the limit of the absolute value of the ratio of consecutive terms in the series. If this limit is less than 1, the series converges, and the radius of convergence is equal to the limit. If the limit is greater than 1, the series diverges, and if it is equal to 1, the ratio test is inconclusive and another test must be used.

## 4. Can the radius of convergence be negative?

No, the radius of convergence must be a positive value. If the radius of convergence is negative, it means the series diverges for all values of x and does not have a valid radius of convergence.

## 5. How is the addition of power series performed?

The addition of power series is done by adding the corresponding coefficients of each term. However, it is important to first check that the series have the same radius of convergence, as adding two series with different radii of convergence can lead to incorrect results.

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