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| Is a summation from n = 1 to infinity of "1" = "1"? |
| Yes. |
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[ 3 ] |
| I don't know. |
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[ 3 ] |
| No. |
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| That doesn't make sense. You must have done something incorrectly. |
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| Total Votes : 10 |
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 Posted: Sun Sep 02, 2007 4:56 pm
Hey,
Dunno how quickly this thread will be read by anyone, but, in case it's before Wednesday:
I'm working on homework posted at http://mtsu.edu/~phys3150/Homework/homework.html
While attempting problem 1.10 #16, I came across something I'd not seen before:
When x = 3,
the sum of "(x-1)^n / (2^n)" from n = 1 to infinity
= the sum of "2^n / 2^n" from n = 1 to infinity
= the sum of "1" from n = 1 to infinity.
Well, that last statement doesn't make sense to me. There is no 'n'! Or do the numerator and denominator not cancel each other?
I said the series was simply = 1 (and was thus convergent). Is that wrong?
Am I to treat it as infinity / infinity, in order to not cancel out the 'n'? I mean, when I try the Ratio Test I get rho = 1, which is inconclusive -- I think because it's not a series at all, simply 1 ...
Does any of what I've said make sense? I mean, do you get what I'm seeing and doing here? Is a summation from n = 1 to infinity of a constant simply that constant?
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Posted: Sun Sep 02, 2007 5:35 pm
Aetherius Lamia Hey, Dunno how quickly this thread will be read by anyone, but, in case it's before Wednesday: I'm working on homework posted at http://mtsu.edu/~phys3150/Homework/homework.html While attempting problem 1.10 #16, I came across something I'd not seen before: When x = 3, the sum of "(x-1)^n / (2^n)" from n = 1 to infinity = the sum of "2^n / 2^n" from n = 1 to infinity = the sum of "1" from n = 1 to infinity. Well, that last statement doesn't make sense to me. There is no 'n'! Or do the numerator and denominator not cancel each other? I said the series was simply = 1 (and was thus convergent). Is that wrong? Am I to treat it as infinity / infinity, in order to not cancel out the 'n'? I mean, when I try the Ratio Test I get rho = 1, which is inconclusive -- I think because it's not a series at all, simply 1 ... Does any of what I've said make sense? I mean, do you get what I'm seeing and doing here? Is a summation from n = 1 to infinity of a constant simply that constant? Haven't done these in a while. Ugh I feel stupid. Wouldn't this be inconclusive by the ratio test? Why is it that when I looked at that problem I immediately thought of Raabe. Sorry if I am not helpful, its been about three years since I last used one these. Hopefully someone will correct me and remind me what I have forgotten.
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Posted: Sun Sep 02, 2007 5:51 pm
Yes, i tried using the ratio test and it is inconclusive; 2^(n+1)/itself = 1, same with 2^n, so you have "1*1" ...
but the question is, what does it mean?
How can you sum "1" from n = 1 to infinity when there is no 'n' in the expression?!
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Posted: Sun Sep 02, 2007 8:26 pm
1^n where n=1 to n=infinity will always equal one. It doesn't matter how many times you multiply 1 by 1 it will still equal 1. therefor the 'n' was left out as it is unnessecary
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Posted: Mon Sep 03, 2007 8:30 am
The way to think about these things is to plug them into a formula. In general, you're looking at  where the big sigma means "sum of a_n for n=1 to infinity". So you simply calculate a_1+a_2+a_3+... So here we have an n in the thing being summed. Now, the question asks for the sum of "1" from n = 1 to infinity. You're supposed to interpret that as saying that a_n = 1 in the above expression for all n. So replacing all of the a_ns by 1 gives you 1+1+1+1+...
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Posted: Mon Sep 03, 2007 8:44 pm
The summation of any constant will be divergent. The equation for your series simplifies to one, which means it is divergent.
As Larya stated, the summation becomes an infinite series of 1 + 1.
Edit: That should be "any non-zero constant."
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Posted: Wed Sep 05, 2007 11:25 am
What I don't understand is that it's "2^n / 2^n" ... which equals 1.
So you're summing 1 as n goes from 1 to infinity, BUT THERE IS NO 'N' IN THE EXPRESSION!
Or is it 1^n as another suggested?
I mean, the expression is -- and this is a power series? --
(x-1)^n / 2^n
so when x = 3, it reduces to 2^n / 2^n ... = 1.
So is it 1+1+1+1+... ? Or is it simply 1, a constant pulled outside the series, so you have 1 * 0+0+0+0+0 ...? Well, I suppose that cannot be correct because 1 * 0 = 0.
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Posted: Wed Sep 05, 2007 5:10 pm
Aetherius Lamia What I don't understand is that it's "2^n / 2^n" ... which equals 1. So you're summing 1 as n goes from 1 to infinity, BUT THERE IS NO 'N' IN THE EXPRESSION! So? What difference does it make if "n" is in the equation? Why would you do anything differently? As already stated, the series is divergent as it infinitely adds one.
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Posted: Fri Sep 07, 2007 7:35 am
When there's no n, then you just say a_n = 1 for all n. If you want, you can write them as 1_1+1_2+1_3... but it's all just 1. The presence of n is just a convenience, a place-holder. After all, when you evaluate, the n gets replaced by an actual number (and thus disappears), so if there's no n to begin with, then the evaluation is much simpler.
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Posted: Tue Sep 11, 2007 6:18 pm
Ah, I think I see ...! I changed my answer to say that it was divergent (continually adding one), and Dr. Montemayor didn't say anything about it ... he said he was only giving them a cursory glance, though, so he could have overlooked it.
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Posted: Tue Sep 11, 2007 10:42 pm
Aetherius Lamia Ah, I think I see ...! I changed my answer to say that it was divergent (continually adding one), and Dr. Montemayor didn't say anything about it ... he said he was only giving them a cursory glance, though, so he could have overlooked it. Well, it is divergent, in the reals at least.
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Posted: Wed Dec 12, 2007 9:34 pm
If you look at the series carefully, you'll see that it is a geometric series.
Sum[((x-1)/2)^n, {n, 1, inf}]
Thus Abs[(x-1)/2] < 1
and so Abs[x-1] < 2
This gives you an interval of convergence of -1 < x < 3
Testing the endpoints,
x = -1
(-1 - 1)/2 = -2/2 = -1
and the sum is Sum[(-1)^n, {n,1,inf}] which is divergent.
From this thread, we've also seen that x = 3 causes the series to diverge.
So (-1, 3) is the correct interval of convergence.
What this means is that as long as x is within the interval, the series will converge. (In fact, it will converge to (x-1)/2 / (1 - (x-1)/2)). Outside of the interval it will diverge.
I know: more than you asked for, but interesting nonetheless.
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