Addition of power series and radius of convergence

[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?

 

[Orodruin]

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]

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?

 

[Orodruin]

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

 

[Orodruin]

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]

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.

 

[Poetria]

Therefore it converges for every x, am I right?

 

[Orodruin]

Yes.

 

[Poetria]

Yes.

Great. :) Many thanks. :)

 

[Ray Vickson]

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]

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.

 

[Ray Vickson]

:) 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.

 

[Orodruin]

In a way, this question is trickier than it looks

I would argue that this is the entire point of the question...

 

[Poetria]

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.