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Topic: find sum (Read 818 times) |
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tony123
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find sum
1/(1.2.3) +1/((4.5.6) + 1/(7.8.9)+......
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jollytall
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Re: find sum
« Reply #1 on: Jun 29th, 2007, 12:02pm » |
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what is 1.2.3 ?
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Sameer
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Re: find sum
« Reply #3 on: Jun 29th, 2007, 12:57pm » |
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If it is the standard notation then it should imply multiplication...
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"Obvious" is the most dangerous word in mathematics.
--Bell, Eric Temple
Proof is an idol before which the mathematician tortures himself.
Sir Arthur Eddington, quoted in Bridges to Infinity
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tony123
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Re: find sum
« Reply #4 on: Jun 29th, 2007, 1:57pm » |
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'.' is a multiplication. 
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username101
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Re: find sum
« Reply #6 on: Jun 30th, 2007, 12:46am » |
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1850? if by the ..... you the patern goes on for one more
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Barukh
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Re: find sum
« Reply #7 on: Jul 2nd, 2007, 8:40am » |
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Is this Medium? 
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towr
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Re: find sum
« Reply #8 on: Jul 2nd, 2007, 10:10am » |
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I'm still not sure what the question is, either there are too many, or too few parenthesis.
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Barukh
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Re: find sum
« Reply #9 on: Jul 3rd, 2007, 3:35am » |
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I think there is an extra parenthesis, so the series in question is:
 1/[ (3n+1)(3n+2)(3n+3) ], n = 0, 1, ...
What about this sum:
1/[ (2n+1)(2n+2) ], n = 0, 1, ...
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pex
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Re: find sum
« Reply #10 on: Jul 3rd, 2007, 4:16am » |
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on Jul 3rd, 2007, 3:35am, Barukh wrote:IWhat about this sum:
1/[ (2n+1)(2n+2) ], n = 0, 1, ...
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That one is much easier; it is ln(2) .
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tony123
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Re: find sum
« Reply #11 on: Jul 3rd, 2007, 4:18am » |
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1/[ (3n+1)(3n+2)(3n+3) ], n = 0, 1, ...
yes i mean this sum
can you help me
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Barukh
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Re: find sum
« Reply #12 on: Jul 3rd, 2007, 5:12am » |
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on Jul 3rd, 2007, 4:16am, pex wrote:| That one is much easier; it is ln(2) . |
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Right! 
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towr
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Re: find sum
« Reply #13 on: Jul 3rd, 2007, 5:22am » |
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Divide and conquer?
(1/2)/(3n+1) -1/(3n+2) + (1/2)/(3n+3) = 1/[(3n+1)(3n+2)(3n+3)]
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| « Last Edit: Jul 3rd, 2007, 5:30am by towr » |
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pex
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Re: find sum
« Reply #14 on: Jul 3rd, 2007, 5:28am » |
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on Jul 3rd, 2007, 5:22am, towr wrote:Divide and conquer?
(1/2)/(3n+1) -1/(3n+2) + (1/2)/(3n+3) = 1/[(3n+1)(3n+2)(3n+3)] |
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I tried that as well, but the asymptotic expressions weren't any easier than the original problem (infinite sums involving zeta(k) multiplied by 3-k, and the like), so I hope someone else can find something easier. (I cannot post the exact results now, because I am at work and my notes are at home.)
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Barukh
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Re: find sum
« Reply #15 on: Jul 3rd, 2007, 6:18am » |
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on Jul 3rd, 2007, 5:28am, pex wrote:| I tried that as well, but the asymptotic expressions weren't any easier than the original problem (infinite sums involving zeta(k) multiplied by 3-k, and the like), so I hope someone else can find something easier. (I cannot post the exact results now, because I am at work and my notes are at home.) |
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Could you please provide some formulas later?
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Sameer
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Re: find sum
« Reply #16 on: Jul 3rd, 2007, 8:55am » |
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I think this is a divergent series.. so sum doesn't exist ... 
The expression reduces to (1/2(3k+1)) - (1/(3k+2)) + (1/2(3k+3))
Applying similar logic to Barukh's alternative question, we get divergence on the final sum.
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Proof is an idol before which the mathematician tortures himself.
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Aryabhatta
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Re: find sum
« Reply #17 on: Jul 3rd, 2007, 9:02am » |
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on Jul 3rd, 2007, 8:55am, Sameer wrote:I think this is a divergent series.. so sum doesn't exist ...
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The nth term 1/(3n+1)(3n+2)(3n+3) < 1/(3n+1)3
So the series is definitely convergent and its value is < Zeta(3)
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towr
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Re: find sum
« Reply #18 on: Jul 3rd, 2007, 10:35am » |
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on Jul 3rd, 2007, 9:02am, Aryabhatta wrote:
The nth term 1/(3n+1)(3n+2)(3n+3) < 1/(3n+1)3
So the series is definitely convergent and its value is < Zeta(3) |
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isn't it also greater than 1/(3n+2)3 ?
That would give an upper and lower bound at least.
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| « Last Edit: Jul 3rd, 2007, 10:36am by towr » |
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Obob
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Re: find sum
« Reply #19 on: Jul 3rd, 2007, 10:51am » |
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I've come up with a cool way of attacking the problem.
Let ak=1/((3k+1)(3k+2)(3k+3)) be the kth term in the series, and define f(x,y,z)= sum akx3k+1y3k+2z3k+3. Holding any of the three variables constant with absolute value <= 1, the resulting function is holomorphic on the unit disc, being bounded by a convergent geometric series. Complex analysis tells us that d3f/dxdydz= sum x3ky3k+1z3k+2, which in turn equals yz2/(1-(xyz)3). So now we have a differential equation d3f/dxdydz = yz2/(1-(xyz)3), and we wish to find f(1,1,1). Now (d2f/dydz)(1,y,z)=integral01 d3f/dxdydz dx + (d2f/dydz)(0,y,z) = integral01 yz^2/(1-(xyz)3) dx. This integral can be explicitly solved in terms of natural logs and arctangents. Next take another integral to remove the derivative with respect to y, and another to remove the derivative with respect to z. All of these integrals could be done by hand. In the end, the final answer is (1/12)(31/2 pi - 3 log 3), which is about .1788.
I think this method works to find any sum of the form 1/(1 . 2. ... . k)+1/((k+1) . (k+2) . ... . (2k))+..., although it is possible that the integrals become unsolvable.
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| « Last Edit: Jul 3rd, 2007, 10:51am by Obob » |
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Sameer
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Re: find sum
« Reply #20 on: Jul 3rd, 2007, 1:50pm » |
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I am not too sure about the subject but I was reading up zeta functions and I came across Hurwitz Zeta Function and series representation by Helmut Hass. Can we use that here?
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"Obvious" is the most dangerous word in mathematics.
--Bell, Eric Temple
Proof is an idol before which the mathematician tortures himself.
Sir Arthur Eddington, quoted in Bridges to Infinity
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SWF
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Re: find sum
« Reply #21 on: Jul 3rd, 2007, 5:33pm » |
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[sum]x3k+k/(3n+k) = [sum][integral] t3n+k-1 dt = [integral] tk-1/(1-t^3) dt
(sum is on n=0 to infinity, integral is on t from 0 to x.
Use Sameer's decomposition of each term into 3 terms of the above form with k=1, k=2, and k=3, and let x approach 1 giving:
(1/2) * [integral] (1-t)/(1+t+t^2) dt (limits on integral are 0 to 1) = [pi]*sqrt(3)/12 -log(3)/4 = 0.1788...
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