https://www8.imperial.ac.uk/content/...ls/M1P1Q17.pdf
I'm stuck on 5. What exactly is wrong with that? The theorem about adding sequences (if a_n converges to a, b_n converges to b then a_n + b_n converges to a+b) is for adding two sequences but I proved by induction that we can add any number of sequences and add the limits. So what exactly is wrong with this in question 5? I think that errors like this wouldn't happen if people wrote out sums properly, rather than using these silly ellipsis marks.
And a question about functions and sequences. Let f(n) be a sequence and let the limit of f(n) be l (where we have f(n) explicitly as a function of n). Define a function on the real numbers by f(x) (so it's a function defined by the same formula as for our sequence). Is the limit as x goes to infinity also l?

gangsta316
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 30012010 00:04

generalebriety
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(Original post by gangsta316)
I proved by induction that we can add any number of sequences and add the limits.
(Original post by gangsta316)
And a question about functions and sequences. Let f(n) be a sequence and let the limit of f(n) be l (where we have f(n) explicitly as a function of n). Define a function on the real numbers by f(x) (so it's a function defined by the same formula as for our sequence). Is the limit as x goes to infinity also l?
* though this is straight out of my head, and there might well be some complicated counterexample in some topological space somewhere, but I'm tired so I can't be bothered to think of one; anyway, this is certainly true if the domain of f is R^n and its range is in R^m, and I don't immediately see why it wouldn't be true for a general nonpathological metric spaceLast edited by generalebriety; 30012010 at 01:55. 
gangsta316
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 30012010 23:07
(Original post by generalebriety)
No you didn't. You proved by induction that we can add any constant number of sequences and add the limits. It's absolutely nothing to do with the ellipsis; while it is true that the terms are all going to zero as n increases, it is also true that the number of terms is increasing as n increases (because there's an n in the limits of the summation). In order to know what the limit is, we need to know which is changing faster! If the terms are all going to zero very quickly but the number of terms is increasing slowly, it'll probably still go to zero; if the terms are all going to zero very slowly but the number of terms is increasing really fast, it might well be that increasing n increases the sum rather than decreasing it.
You haven't defined f on the real numbers, you've only defined it on the natural numbers. If f(x) converges as x tends to infinity, then yes, it must tend to l. You might have a theorem in your notes telling you something along the lines of "f(x) > l as x > a if and only if f(x_n) > l as n > infinity for every sequence x_n > a which doesn't leave the domain of f"*  this also holds true when a is infinity, and the proof is similar. The "if" implication only holds if f(x_n) converges for all such x_n, though  for example, suppose f(n) = 0 for all n; we might have that f(x) = 0 identically (which converges), or that f(x) = sin(pi*x) (which diverges).
* though this is straight out of my head, and there might well be some complicated counterexample in some topological space somewhere, but I'm tired so I can't be bothered to think of one; anyway, this is certainly true if the domain of f is R^n and its range is in R^m, and I don't immediately see why it wouldn't be true for a general nonpathological metric space
I still don't fully understand. I proved that we can add a (natural) number of sequences and the limit of the sum is the sum of the limits. Why does this become invalid just because we are varying the number of sequences? 
DFranklin
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 31012010 02:00
GE's already said it pretty well.
When you prove a_n+b_n > a+b, the proof is basically saying that the "error" terms go to zero as n goes to infinity, and as the sum of two things that go to zero is zero, the total error goes to zero as well.
But if you get the nth term of the sequence by summing 'n' sequences (e.g. as in the question), then knowing that the error for each sequence goes to 0 isn't enough, because you have n such errors to add together, and it's perfectly possible for the sum of n terms to *not* go to 0.
If you specifically want to know why your proof doesn't apply: you've proved that for any +ve integer N, you can sum N sequences and the limit will be the sum of the individual limits. But for the example, however you choose N, you will eventually be summing more than N sequences. And you certainly didn't prove that you could sum an infinite number of sequences and the limit would be the sum of the individual limits. 
generalebriety
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 31012010 17:30
(Original post by gangsta316)
For the first bit:
I still don't fully understand. I proved that we can add a (natural) number of sequences and the limit of the sum is the sum of the limits. Why does this become invalid just because we are varying the number of sequences?
Here's a bit more of a sensible heuristic argument. Take some sequences that are all tending to zero. Then of course any one of them eventually tends to zero; the sum of any two of them eventually tends to zero; the sum of any three of them eventually tends to zero; etc. This is easy to show: if you're adding any fixed number n of them together, you just need to pick arbitrary epsilon > 0 as normal, wait until each sequence is < epsilon/n, and then you've obviously got the sum < epsilon. But this requires n to be constant! If n is varying, then you won't necessarily get each sequence < epsilon/n, because epsilon/n isn't fixed either. Now, if epsilon/n is decreasing rather quickly and some of the sequences are tending to zero rather slowly, then their sum might always remain greater than epsilon/n, and in particular might not converge.
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