- #1

nobahar

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**Limit of otherwise equal summations is uneuqal due to "indeterminate difference"?**

Hello!

I was working through a section on summation in a maths textbook because of a recent question I asked on the forum, and I came across something which is confusing. I am not sure if it is necessary, but here is the proof first of all:

If

[tex]\lim_{n\to{\infty}}\left(\sum_{i=1}^n a_{i}\right) = A[/tex]

then the sum converges and [itex]a_{n}\to0[/itex] as [itex]n\to{\infty}[/itex].

It goes:

If [itex]S_{n} = a_{1} + a_{2} + a_{3} + … + a_{n}[/itex] then

[tex]S_{n} – S_{n-1} = \sum_{i=1}^n a_{i} - \sum_{i=1}^{n-1} a_{i} = a_{n}[/tex]

Taking the limit gives the required result:

[tex]\lim_{n\to{\infty}}(a_{n}) = \lim_{n\to{\infty}}(S_{n} –S_{n-1}) = \lim_{n\to{\infty}}(S_{n}) - \lim_{n\to{\infty}}(S_{n-1}) = A – A = 0[/tex]

I tried the proof myself first, and instead tried substituting [itex]S_{n} – S_{n-1}[/itex] in place of [itex]a_{i}[/itex] in [itex]\sum_{i=1}^n a_{i}[/itex].

This gives:

[tex]\sum_{i=1}^n (S_{n} – S_{n-1}) = \sum_{i=1}^n \left(\sum_{j=1}^i a_{j} - \sum_{i=1}^{i-1} a_j\right)[/tex]

If I take the limit at this point, it is my understanding that this is still:

[tex]a_{1} + a_{2} + a_{3} + … = A[/tex]

However:

[tex]\sum_{i=1}^n \left(\sum_{j=1}^i a_{j} - \sum_{i=1}^{i-1} a_j\right) = \sum_{i=1}^n \sum_{j=1}^i a_{j} - \sum_{i=1}^n \sum_{i=1}^{i-1} a_j[/tex]

And I think in the limit the difference between i and i-1 disappears:

[tex]\lim_{n\to{\infty}}\left(\sum_{i=1}^n \sum_{j=1}^i a_{j}\right) = \lim_{n\to{\infty}}\left(\sum_{i=1}^n \sum_{i=1}^{i-1} a_j\right)[/tex]

This is where I’m confused. If:

[tex]\lim_{n\to{\infty}}\left(\sum_{i=1}^n \sum_{j=1}^i a_{j}\right)[/tex]

was an actual number, say B, then

[tex]\lim_{n\to{\infty}}\left(\sum_{i=1}^n \sum_{j=1}^i a_{j} - \sum_{i=1}^n \sum_{i=1}^{i-1} a_j\right) = B - B = 0[/tex]

And the equality in the limit for the following equation (equation *):

[tex]\lim_{n\to{\infty}}\left(\sum_{i=1}^n \left(\sum_{j=1}^i a_{j} - \sum_{i=1}^{i-1} a_j\right)\right) = \lim_{n\to{\infty}}\left(\sum_{i=1}^n \sum_{j=1}^i a_{j} - \sum_{i=1}^n \sum_{i=1}^{i-1} a_j\right)[/tex]

doesn’t make sense (even though it is otherwise true), since

[tex]\lim_{n\to{\infty}}\left(\sum_{i=1}^n \left(\sum_{j=1}^i a_{j} - \sum_{i=1}^{i-1} a_j\right)\right) = \lim_{n\to{\infty}}\left(\sum_{i=1}^n (S_{n} – S_{n-1})\right) = A[/tex]

That is, the lim on the LHS of equation * is A, whereas on the RHS it is 0 (if it produced an actual number, B, in the limit).

I figured that the reason the equality is otherwise true but not in the limit is because the RHS produces an indeterminate form, which I think is called the "indeterminate difference", and so the result doesn't mean anything (like 0/0 or infty/0, etc):

[tex]\lim_{n\to{\infty}}\left(\sum_{i=1}^n \sum_{j=1}^i a_{j}\right)\to{\infty}[/tex]

Since:

[tex]\sum_{i=1}^n \sum_{j=1}^i a_{j} = (a_{1}) + (a_{1} + a_{2}) + (a_{1} + a_{2} + a_{3}) + … + (a_{1} + a_{2} + a_{3} + … + a_{n}) = na_{1} + (n-1)a_{2} + (n-2)a_{3} + … + a_{n}[/tex]

Then:

[tex]\lim_{n\to{\infty}}\left(\sum_{i=1}^n \sum_{j=1}^i a_{j}\right) = \lim_{n\to{\infty}}\left(na_{1} + (n-1)a_{2} + (n-2)a_{3} + … + a_{n}\right)[/tex] which [itex]\rightarrow {\infty}[/itex]

Therefore:

[tex]\lim_{n\to{\infty}}\left(\sum_{i=1}^n \sum_{j=1}^i a_{j} - \sum_{i=1}^n \sum_{i=1}^{i-1} a_j \right) = “{\infty} - {\infty}”[/tex]

Which, form my understanding, doesn’t mean anything, and you have to manipulate the summation to get it into a form where it does make sense. Such as:

[tex]\sum_{i=1}^n \left(\sum_{j=1}^i a_{j} - \sum_{i=1}^{i-1} a_j)\right)[/tex]

Is this somewhat similar to what happens with L’Hopital’s rule, where there’s some “indeterminate form”, so no solution appears to exist when left in that form, but there is an actual solution (in the case of L'Hopital's rule, by using the derivatives)?

Any clarification appreciated, is my understanding anywhere near the mark? ...this took a long time to write!

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