Limits of infinite sums of sequences

In summary: If you adopt a -1* infinity =0 convention then the first sum is greater than the second and the second sum is less than the first.
  • #1
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I understand that the limit of the sum of two sequences equals the sum of the sequences' limis: [tex]\displaysyle \lim_{n\rightarrow\infty} (a_{n} + b_{n}) = \lim_{n\rightarrow\infty}a_{n} + \lim_{n\rightarrow\infty}b_{n}[/tex]. Similar results consequenly hold for sums of three sequences, four sequences, etc. (Call this the "finite distributive property.") But what about the sum of countably many sequences? Given the sequences [tex] \{a_{i} (n)\} [/tex], is it the case that

[tex]\displaystyle \lim_{n\rightarrow\infty} \sum_{i=1}^{\infty} a_{i}(n) = \sum_{i=1}^{\infty} \lim_{n\rightarrow\infty}a_{i}(n) ?[/tex]

If not, are there any extra conditions on the sequences sufficient to make the equality hold? Any necessary and sufficient ones? I have an argument that if if both sides of the equality are finite, then the it holds. However, I've very likely made a mistake and would appreciate it if you guys could/would point out any errors.

Let [tex]\displaystyle L := \Sigma_{i=1}^{\infty} \lim_{n\rightarrow\infty}a_{i}(n)[/tex]. So [tex]\displaysyle \forall \epsilon > 0 \exists S \forall s \geq S \ |L - \Sigma_{i=1}^{s}\lim_{n\rightarrow\infty}a_{i}(n)| < \frac{\epsilon}{2}.[/tex] Also let [tex]L_{n} := \Sigma_{i=1}^{\infty}a_{i}(n)[/tex]. We now therefore have, given the same [tex]\frac{\epsilon}{2}[/tex] as before, [tex]\forall T \exists t \geq T \ |L_{n} - \Sigma_{i=1}^{t} a_{i}(n) | < \frac{\epsilon}{2}[/tex]. Taking n to the limit on both sides yields [tex]\displaystyle \lim_{n\rightarrow\infty} |L_{n} - \Sigma_{i=1}^{t} a_{i}(n) | = |\lim_{n\rightarrow\infty}L_{n} - \Sigma_{i=1}^{t} \lim_{n\rightarrow\infty} a_{i}(n) | \leq \frac{\epsilon}{2}[/tex] (by the finite distributive property). Finally, define [tex]L_{0} := \lim_{n\rightarrow\infty}L_{n}, \ U := \max(S, T)[/tex]. Then for each [tex] u \geq U [/tex], we have both [tex] |L - \Sigma_{i=1}^{u} \lim_{n\rightarrow\infty} a_{i}(n) | < \frac{\epsilon}{2}[/tex] and [tex]|L_{0} - \Sigma_{i=1}^{u} \lim_{n\rightarrow\infty} a_{i}(n) | \leq \frac{\epsilon}{2}[/tex]. By the triangle inequality, then, [tex] |L - L_{0}| < \epsilon[/tex]. Since epsilon was arbitrary, [tex]L = L_{0}[/tex].

What about the cases where the sides aren't finite?
 
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  • #2
I don't know the answer off the top of my head, but I do want to point this out.

What you have is essentially TWO limits. The infinite sum cannot be directly evaluated and so it is actually a limit in disguise. So your question boils down to when can you interchange two limit operations?

I'm not a good analysis student, though, and I don't know the exact conditions.
 
  • #3
necessary and sufficient conditions are hard to come by
uniform convergence is a useful though limited sufficient condition
 
  • #4
O.K., here's an easy counterexample to the general claim:

[tex]\displaystyle \sum_{i=1}^{\infty} \lim_{n\rightarrow\infty} \frac{1}{2^{n}} = 0, \ \lim_{n\rightarrow\infty} \sum_{i=1}^{\infty} \frac{1}{2^{n}} = \infty. [/tex]

Limits usually behave so naturally in terms of commuting with other operators that I'm still somewhat surprised they don't commute here.
 
  • #5
Limits usually behave so naturally in terms of not commuting with other operators that I'm still somewhat unsurprised they don't commute here. Often uniform situations are used that gives commutation. It is easy to produce counter examples by using nonuniform limits.
 
  • #6
Hey mag487,

Did you get to verify if your argument was correct? I went through your proof and couldn't find any loophole; however I am no expert in analysis.
 
  • #7
mag487 said:
O.K., here's an easy counterexample to the general claim:

[tex]\displaystyle \sum_{i=1}^{\infty} \lim_{n\rightarrow\infty} \frac{1}{2^{n}} = 0, \ \lim_{n\rightarrow\infty} \sum_{i=1}^{\infty} \frac{1}{2^{n}} = \infty. [/tex]

Limits usually behave so naturally in terms of commuting with other operators that I'm still somewhat surprised they don't commute here.

Since the 1/2^n doesn't have an i in it you can take it out of the summation in both cases so you get 0*infinity in both cases. If you adopt a 0* infinity =0 convention then both sides are equal to zero.
 

1. What is the definition of a limit of an infinite sum of sequences?

The limit of an infinite sum of sequences, also known as a limit of an infinite series, is the value that the sum approaches as the number of terms in the sequence increases towards infinity.

2. How do you determine if an infinite sum of sequences has a limit?

To determine if an infinite sum of sequences has a limit, you can use various methods such as the ratio test, the root test, or the comparison test. These tests help determine if the terms in the sequence decrease or increase towards a finite value, indicating the existence of a limit.

3. Can an infinite sum of sequences have more than one limit?

No, an infinite sum of sequences can only have one limit. If the limit does not exist, it means that the sum does not approach a finite value and instead diverges to infinity or negative infinity.

4. What is the role of convergence in determining the limit of an infinite sum of sequences?

Convergence is an essential concept in determining the limit of an infinite sum of sequences. If a sequence is convergent, it means that the terms in the sequence approach a finite value, indicating the existence of a limit. On the other hand, if a sequence is divergent, the limit does not exist.

5. Can an infinite sum of sequences with negative terms have a limit?

Yes, an infinite sum of sequences with negative terms can have a limit, as long as the terms decrease towards a finite value. The limit, in this case, would be a negative number, indicating the sum approaches a negative value as the number of terms increases towards infinity.

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