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Topic: Sum to infinity of 1/(1+n^2) (Read 4331 times) |
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Aryabhatta
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Re: Sum to infinity of 1/(1+n^2)
« Reply #1 on: Aug 25th, 2004, 9:30am » |
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My first guess is : pi/2, by considering the integral of 1/(1+x2) in 0 to [infty]
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TenaliRaman
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Re: Sum to infinity of 1/(1+n^2)
« Reply #2 on: Aug 25th, 2004, 9:32am » |
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(pi*coth(pi)-1)/2
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| « Last Edit: Aug 25th, 2004, 9:58am by TenaliRaman » |
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NickH
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Re: Sum to infinity of 1/(1+n^2)
« Reply #3 on: Aug 26th, 2004, 1:39pm » |
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That's rather cryptic, TenaliRaman! What was your method?
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TenaliRaman
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Re: Sum to infinity of 1/(1+n^2)
« Reply #4 on: Aug 27th, 2004, 7:36am » |
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f(z) = (pi*cot(pi*z))/(z^2+1)
Contour Integral of f(z)
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Smitty Worbenmenjenson
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Re: Sum to infinity of 1/(1+n^2)
« Reply #5 on: Jan 23rd, 2005, 3:19pm » |
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Set the Limit of the Sum to Infinity and you will find that 1/infinity gets really small (and insignificant); in fact it is close to 0.
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Icarus
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Re: Sum to infinity of 1/(1+n^2)
« Reply #6 on: Jan 23rd, 2005, 8:36pm » |
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If 1/[infty] is to be given a value at all, then 1/[infty] is equal to 0, not just close to it.
But infinite sums like this do not involve an actual infinity. Rather, the sum is defined to be the unique value S to which you can get as close to as you want (shy of actually hitting it) by sufficiently many terms.
To provide an example, look at the much simpler sum
[sum]n=0[supinfty] 1/2n = 1 + 1/2 + 1/4 + 1/8 +...
It is easy to see algebraically that
[sum]n=0N 1/2n = 2 -1/2N.
So if I would like to get within 0.01 of 2, I note that 1/27 [approx] 0.0078 < .01, So if I sum terms n=0 through n=7, I can get this close to 2. If I want to get within 0.001 of 2, I have to sum terms n=0 through n=10, etc. No matter how small a margin is set, I can always get within that distance of 2 by adding up enough terms. And adding more terms after will not take me farther away, either.
Note also that 2 is the only number with this property. The sums are all < 2, so for any x > 2, I can't get any closer than x - 2 to x. As for x < 2, a few I can hit dead on, but only by stopping there. If I add in any more terms, the sum moves away, and never returns.
For these reasons, if we actually could add up an infinite number of terms, the result could only be 2. So we define the infinite sum to be 2.
The value of any other infinite sum, such as the one in this problem, is defined in a similar manner. You find the unique number that is approached as closely as you can want, without the sum moving away again when more terms are added. If such a number exists, the sum is said to "converge" to that value. If no such number exists (as with, for example, [sum]n=0[supinfty] 1), the sum is said to "diverge", and no value is defined for it.
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