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11/2+1/31/41/5.... I solve and get 0. amazing. Where is it wrong? 
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#1
Feb2612, 04:32 AM

P: 318

11/2+1/31/4+1/51/6+1/7........ ∞
let take two series S1 and S2 S_{1}=1+1/3+1/5+1/7....... and S_{2} =1/2+1/4+1/6+1/8........... we are intended to find out S_{1} S_{2}. 2S_{2}=1+1/2+1/3+1/4...... S_{1}+S_{2}=1+1/2+1/3+1/4+1/5+1/6+1/7........ So 2S_{2}=S_{1}+S_{2} => S_{1}=S_{2}.... \/\/\/\/\/\/\/ =>S_{1}S_{2}=0? /\/\/\/\/\/\/\ Amazing . Afterall it's incorrect since every odd place number is greater then even place number so it should positive.. 


#2
Feb2612, 05:03 AM

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The first strict mathematical flaw is that S_{1} and S_{2} are not numbers (the series sum to infinity).
On a more subtle level, it is in fact possible to rearrange the infinite alternating series S to get any number you desire  this is true of any series which is convergent, but not absolutely convergent. Are you familiar with the concepts of convergent/divergent series? 


#3
Feb2612, 10:29 AM

P: 318

S_{1}and S_{2} are decreasing HP(?). Does they tends to infinity?? I have learnt about infinite GP. That sum to a constant number. Finally can u tell me how to calculate the sum of the required series. 


#4
Feb2612, 12:32 PM

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11/2+1/31/41/5.... I solve and get 0. amazing. Where is it wrong?
So how to evaluate the original series? What that series is not absolutely convergent, it is convergent in the sense that the partial sums converge to a definite finite value. As for how to evaluate the original series, suppose you write the auxiliary series [tex]S'(x) = 1*x1/2*x^2+1/3*x^31/4*x^4+\cdots = \sum_{n=1}^{\infty}\frac{(1)^{n+1}x^n}{n}[/tex] What is this series S'(x)? (i.e., what function does it represent?) What happens at x=1? 


#5
Feb2612, 12:51 PM

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The fact that you are asking this question when you have ''learned about infinite geometric progressions'' is good, because it shows you are interested in math and thinking about what you are doing, and not just learning how to pass the next test.
There are lots of areas of math where "simple" questions turn out to be hard to answer. The sum is actually ##\log_e 2## = approximately 0.6931, but you need to learn calculus to understand why that is the answer. If you want to try to find the sum by hand (or with a computer), then as post #2 said, you have to add the numbers up in the same order as the original series. If you rearrange them, you can get more or less any answer you like. You can get two estimates that are too big and too small by taking the terms in pairs. 1  1/2 = 1/(1.2) 1/3  1/4 = 1/(3.4) 1/5  1/6 = 1/(5.6) etc So the sum = 1/2 + 1/12 + 1/30 + ... Or, take the first term on its own and the rest in pairs. 1/2 + 1/3 = 1/(2.3) 1/4 + 1/5 = 1/(4.5) 1/6 + 1/7 = 1/(6.7) etc So the sum = 1  1/6  1/20  1/42 .... That will give you two estimates that bracket the answer, but this series is very slow to converge, so you will have to take hundreds of terms to find the answer to a few decimal places. 


#6
Feb2612, 01:28 PM

P: 318

adding them like simple sum is dummy style. getting sum of this series is part of question in physics. So i must have been read that in mathematics but not getting that. this is part of an electrostatics question where i am required to find potential at origin when negative charges are placed at odd positions (on x axis) and positive charges are placed at even position. You say i need to know calculus for it. i think i know. Please see my syllabus(calculus is at lower portion) and give me a proper hint to approach this question. 


#7
Feb2612, 06:16 PM

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#8
Feb2612, 06:26 PM

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If you know how to write functions like ##e^x##, ##\sin x##, or ##\log(1+x)## as power series, then it should be obvious why the answer is ##\log 2## (see D.H's post) but you haven't covered that, I don't know any other way to find the limit. 


#9
Feb2612, 09:12 PM

P: 318

is it know possible to solve it. thanks for your replies. 


#10
Feb2612, 09:15 PM

P: 318

what is this divergent and convergent series???



#11
Feb2712, 05:56 AM

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P: 15,203

A series is convergent if the sequence of its partial sums converges to a finite value. If the sequence of partial sums does not converge to a finite value the series is divergent. For example, the series 11+11+11+... (Grandi's series) is divergent because the sequence of partial sums alternates between 1 and 0. The series 1+1/2+1/3+1/4+... (the harmonic series) is divergent because the sequence of partial sums grows without bound.
Note that your series, 11/2+1/31/4+... differs from the harmonic series only in the signs of the terms. Your series is called the alternating harmonic series. There's a fairly simple test for convergence for alternating series (series whose elements alternate between positive and negative). Such a series is convergent if the sequence comprising the absolute values of the terms the series is monotonically decreasing and converges to zero as n approaches infinity. Your series passes this test and hence will converge to some finite value. Compare your series to 11/2+1/41/8+... While the tricks that you tried to use on your series won't work on that series, they will work on this one. The reason is that the corresponding series 1+1/2+1/4+1/8+... is also convergent. Series such as 11/2+1/41/8+... are called absolutely convergent series. Rearrange the terms of an absolutely convergent series and you always get the same sum. Different arrangements of the terms of a conditionally convergent series (one that is not absolutely convergent) can yield different sums. In fact, you can rearrange the terms of a conditionally convergent series to give any sum you want. 


#12
Feb2812, 12:03 AM

P: 99

No, [tex]e = \displaystyle\lim_{x\to 0}{(1 + x)}^{\frac{1}{x}}[/tex] [tex]e^x = 1 + \dfrac{x}{1!} + \dfrac{x^2}{2!} + \dfrac{x^3}{3!} + \ ...[/tex] 


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