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

1. Aug 25, 2012

nobahar

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
$$\lim_{n\to{\infty}}\left(\sum_{i=1}^n a_{i}\right) = A$$
then the sum converges and $a_{n}\to0$ as $n\to{\infty}$.

It goes:

If $S_{n} = a_{1} + a_{2} + a_{3} + … + a_{n}$ then
$$S_{n} – S_{n-1} = \sum_{i=1}^n a_{i} - \sum_{i=1}^{n-1} a_{i} = a_{n}$$
Taking the limit gives the required result:
$$\lim_{n\to{\infty}}(a_{n}) = \lim_{n\to{\infty}}(S_{n} –S_{n-1}) = \lim_{n\to{\infty}}(S_{n}) - \lim_{n\to{\infty}}(S_{n-1}) = A – A = 0$$

I tried the proof myself first, and instead tried substituting $S_{n} – S_{n-1}$ in place of $a_{i}$ in $\sum_{i=1}^n a_{i}$.
This gives:
$$\sum_{i=1}^n (S_{n} – S_{n-1}) = \sum_{i=1}^n \left(\sum_{j=1}^i a_{j} - \sum_{i=1}^{i-1} a_j\right)$$
If I take the limit at this point, it is my understanding that this is still:
$$a_{1} + a_{2} + a_{3} + … = A$$
However:
$$\sum_{i=1}^n \left(\sum_{j=1}^i a_{j} - \sum_{i=1}^{i-1} a_j\right) = \sum_{i=1}^n \sum_{j=1}^i a_{j} - \sum_{i=1}^n \sum_{i=1}^{i-1} a_j$$
And I think in the limit the difference between i and i-1 disappears:
$$\lim_{n\to{\infty}}\left(\sum_{i=1}^n \sum_{j=1}^i a_{j}\right) = \lim_{n\to{\infty}}\left(\sum_{i=1}^n \sum_{i=1}^{i-1} a_j\right)$$

This is where I’m confused. If:
$$\lim_{n\to{\infty}}\left(\sum_{i=1}^n \sum_{j=1}^i a_{j}\right)$$
was an actual number, say B, then
$$\lim_{n\to{\infty}}\left(\sum_{i=1}^n \sum_{j=1}^i a_{j} - \sum_{i=1}^n \sum_{i=1}^{i-1} a_j\right) = B - B = 0$$
And the equality in the limit for the following equation (equation *):
$$\lim_{n\to{\infty}}\left(\sum_{i=1}^n \left(\sum_{j=1}^i a_{j} - \sum_{i=1}^{i-1} a_j\right)\right) = \lim_{n\to{\infty}}\left(\sum_{i=1}^n \sum_{j=1}^i a_{j} - \sum_{i=1}^n \sum_{i=1}^{i-1} a_j\right)$$
doesn’t make sense (even though it is otherwise true), since
$$\lim_{n\to{\infty}}\left(\sum_{i=1}^n \left(\sum_{j=1}^i a_{j} - \sum_{i=1}^{i-1} a_j\right)\right) = \lim_{n\to{\infty}}\left(\sum_{i=1}^n (S_{n} – S_{n-1})\right) = A$$
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):
$$\lim_{n\to{\infty}}\left(\sum_{i=1}^n \sum_{j=1}^i a_{j}\right)\to{\infty}$$
Since:
$$\sum_{i=1}^n \sum_{j=1}^i a_{j} = (a_{1}) + (a_{1} + a_{2}) + (a_{1} + a_{2} + a_{3}) + … + (a_{1} + a_{2} + a_{3} + … + a_{n}) = na_{1} + (n-1)a_{2} + (n-2)a_{3} + … + a_{n}$$
Then:
$$\lim_{n\to{\infty}}\left(\sum_{i=1}^n \sum_{j=1}^i a_{j}\right) = \lim_{n\to{\infty}}\left(na_{1} + (n-1)a_{2} + (n-2)a_{3} + … + a_{n}\right)$$ which $\rightarrow {\infty}$

Therefore:
$$\lim_{n\to{\infty}}\left(\sum_{i=1}^n \sum_{j=1}^i a_{j} - \sum_{i=1}^n \sum_{i=1}^{i-1} a_j \right) = “{\infty} - {\infty}”$$
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:
$$\sum_{i=1}^n \left(\sum_{j=1}^i a_{j} - \sum_{i=1}^{i-1} a_j)\right)$$

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!

Last edited: Aug 25, 2012
2. Aug 27, 2012

elvishatcher

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

I don't understand the part where you replace $$a_i$$ with $$S_n - S_{n-1}$$ and then put that into the original sum. It seems to me like you're doing something meaningless here as the sum takes values of i from 1 to n and you are plugging in $$S_n - s_{n-1}$$ That is, you're taking the sum of that expression from i = 1 to i = n but there are no i's in it. I feel like perhaps you confused $$a_n$$ with $$a_i$$ since $$a_n = S_n - S_{n-1}$$ Or am I just understanding you incorrectly?

3. Aug 28, 2012

nobahar

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

Thanks for the response elvishatcher, and you're correct; my apologies. I meant $S_{i}-S_{i-1} = a_{i}$ and then I use the LHS for $\sum_{i=1}^n a_{i}$ to get:
$$\sum_{i=1}^n \left(S_{i} - S_{i-1}\right)$$.
Do you see what I mean towards the end?, the two summations:
$$\sum_{i=1}^n \left(\sum_{j=1}^i a_{j} - \sum_{i=1}^{i-1} a_j\right) = \sum_{i=1}^n \sum_{j=1}^i a_{j} - \sum_{i=1}^n \sum_{i=1}^{i-1} a_j$$
Are equal for any n, except in the limit; i.e. following is not true:
$$\lim_{n\to{\infty}}\left(\sum_{i=1}^n \left(\sum_{j=1}^i a_{j} - \sum_{i=1}^{i-1} a_j\right)\right) = \lim_{n\to{\infty}}\left(\sum_{i=1}^n \sum_{j=1}^i a_{j} - \sum_{i=1}^n \sum_{i=1}^{i-1} a_j\right)$$
Since one gives:
$$\lim_{n\to{\infty}}\left(\sum_{i=1}^n \left(\sum_{j=1}^i a_{j} - \sum_{i=1}^{i-1} a_j\right)\right) = \lim_{n\to{\infty}}\left(\sum_{i=1}^n (S_{n} – S_{n-1})\right) = A$$
And the other gives:
$$\lim_{n\to{\infty}}\left(\sum_{i=1}^n \sum_{j=1}^i a_{j} - \sum_{i=1}^n \sum_{i=1}^{i-1} a_j \right) = “{\infty} - {\infty}”$$
From my attempt at least. I tried to justify these results in the first post, I'm not sure if they are correct. I figured the 'inconsistency' was due to the latter being an "indtereminate difference" (I think that is what it's called), and therefore isn't saying anything that makes sese, in a similar way to 0/0, etc.

Last edited: Aug 28, 2012
4. Aug 28, 2012

elvishatcher

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

Ah, that makes a lot more sense. I'm no expert, but it seems to me that you're correct about the problem being due to ∞ - ∞ not being a defined value. It seems like what you do to get the two different results is solid. Infinity's a kinda weird thing to deal with, and a lot of its complexities are beyond my current knowledge; but the 4th post down in this thread: https://www.physicsforums.com/showthread.php?t=162517 seems to say that you're right about ∞ - ∞ being an undefined/indeterminate/whatever (i'm not sure of the exact term that's appropriate) value, so it's not just equal to 0. That thread might be able to help you out a little more.

5. Aug 28, 2012

SammyS

Staff Emeritus
Re: Limit of otherwise equal summations is uneuqal due to "indeterminate difference"?

nobahar,

Whatever do you mean by $\displaystyle \sum_{i=1}^{i-1} a_j \,,$ or did you mean to write $\displaystyle \sum_{j=1}^{i-1} a_j \ ?$

By the way, ∞ - ∞ is an indeterminate form. Depending upon the specific details of such a limit, a limit which is of that form may turn out to be anything, including possibly ∞ or -∞, or it might be that the limit does not exist.

6. Aug 29, 2012

nobahar

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

Thanks for the link, it does address the question. It's reassuring that someone agrees with the method I used to get two different results, as $\infty$ seems like a tricky thing to deal with. Many thanks for the response.

Damn it! I did mean $\displaystyle \sum_{j=1}^{i-1} a_j$. That's two notation errors... I hope thats it. I copied and pasted where I could to save time (because it took ages and I am not well versed in writing in LaTex) and the error got carried over; I can't believe I didn't notice it, I should have used a letter that didn't look similar to i! Anyway, if I have understood what your saying, the $\infty - \infty$ form essentially says nothing - it doesn't say what the actual limit is or if it even exists: it gives no indication of what the 'true value' is, which "may turn out to be anything". Again, if I've understood correctly, you are in agreement with elvishatcher.
Thanks also for the response.

Last edited: Aug 29, 2012