1-1/2+1/3-1/4-1/5 I solve and get 0. amazing. Where is it wrong?

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In summary, the conversation is about a series of numbers that do not converge to a definite finite value. The concept of convergent/divergent series is discussed and it is explained that the original series is not absolutely convergent. The conversation also touches on how to evaluate the sum of the series, with a hint to use calculus and the mention of the PSLQ algorithm. The conversation ends with a clarification that the question is related to an electrostatics problem.
  • #1
vkash
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1-1/2+1/3-1/4-1/5... I solve and get 0. amazing. Where is it wrong?

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


The first strict mathematical flaw is that S1 and S2 are not numbers (the series sum to infinity).

On a more subtle level, it is in fact possible to re-arrange 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


Office_Shredder said:
The first strict mathematical flaw is that S1 and S2 are not numbers (the series sum to infinity).

On a more subtle level, it is in fact possible to re-arrange 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?
But if i want to calculate the sum of series then what should i do.
S1and S2 are decreasing HP(?). Does they tends to infinity?? I have learned about infinite GP. That sum to a constant number.
Finally can u tell me how to calculate the sum of the required series.
 
  • #4


vkash said:
1-1/2+1/3-1/4+1/5-1/6+1/7... ∞
let take two series S1 and S2
S1=1+1/3+1/5+1/7...
and S2 =1/2+1/4+1/6+1/8...
we are intended to find out S1 -S2.
You can't do that. Those are divergent series.

2S2=1+1/2+1/3+1/4...
You can't do this with a divergent series.

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*x-1/2*x^2+1/3*x^3-1/4*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


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.
 
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  • #6
AlephZero said:
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.

How you got the answer ln2. that's exactly the answer of this question.
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 http://jee.iitd.ac.in/mathematics.php(calculus is at lower portion) and give me a proper hint to approach this question.
 
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  • #7


vkash said:
How you got the answer ln2. that's exactly the answer of this question.

Hint: it's a Taylor series question.

adding them like simple sum is dummy style.

You can call it "dummy style" but in the modern world that is how it's done: do a high precision calculation and use the PSLQ algorithm to try to identify it.
 
  • #8
vkash said:
I have learned about infinite GP. That sum to a constant number.
Finally can u tell me how to calculate the sum of the required series.

I interpreted that as meaning you had learned about infinite geometric progressions and you were trying to use the same ideas to sum this series, but from your question you hadn't done any courses that explained why your idea didn't work for this series. Apologies for that misunderstanding.

vkash said:
You say i need to know calculus for it. i think i know. Please see my http://jee.iitd.ac.in/mathematics.php(calculus is at lower portion) and give me a proper hint to approach this question.

Hmm... I would have said "Taylor series", but that isn't mentioned in your syllabus. That might just be something that wasn't mentioned, but it also says "derivatives up to order two" which suggests your syllabus doesn't cover this topic.

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


AlephZero said:
I interpreted that as meaning you had learned about infinite geometric progressions and you were trying to use the same ideas to sum this series, but from your question you hadn't done any courses that explained why your idea didn't work for this series. Apologies for that misunderstanding.



Hmm... I would have said "Taylor series", but that isn't mentioned in your syllabus. That might just be something that wasn't mentioned, but it also says "derivatives up to order two" which suggests your syllabus doesn't cover this topic.

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.
I know only about ##e^x## it was in limits. ##e^x##= ##limx->0 (1+x)^(1/x)## and on expanding it with binomial you can get a series that is 1+x/1+x^2/2!.... and so on till ∞. that's all i know about e.
is it know possible to solve it.
thanks for your replies.
 
  • #10


what is this divergent and convergent series?
 
  • #11


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 1-1+1-1+1-1+... (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, 1-1/2+1/3-1/4+... 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 1-1/2+1/4-1/8+... 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 1-1/2+1/4-1/8+... 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


vkash said:
I know only about ##e^x## it was in limits. > > ##e^x##= < < ##limx->0 (1+x)^(1/x)## and on expanding it
with binomial you can get a series that is 1+x/1+x^2/2!.... and so on till ∞.
thats all i know about e.


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]
 

1. How did you solve the equation 1-1/2+1/3-1/4-1/5?

The equation was solved using the order of operations: first, the fractions were simplified, then the addition and subtraction were performed from left to right.

2. Why does the solution appear to be 0?

The solution is 0 because when fractions with the same denominator are added or subtracted, the denominator remains the same and only the numerators are added or subtracted. In this case, all the numerators cancel each other out, leaving 0 as the final result.

3. Is there a mistake in the equation or solution?

No, there is no mistake. The solution is 0, which is the correct answer for this equation.

4. Can you explain why the solution is amazing?

The solution may seem amazing because it is counterintuitive to think that adding and subtracting fractions can result in 0. However, it is a common occurrence when the fractions have the same denominator and their numerators cancel each other out.

5. Where could someone go wrong when solving this equation?

One possible mistake someone could make is not simplifying the fractions before performing the addition and subtraction. Another mistake could be performing the operations in the wrong order, such as subtracting before adding or starting from the right instead of the left.

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