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

1. Feb 26, 2012

### vkash

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..

2. Feb 26, 2012

### Office_Shredder

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

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. Feb 26, 2012

### vkash

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

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 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. Feb 26, 2012

### D H

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

You can't do that. Those are divergent series.

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
$$S'(x) = 1*x-1/2*x^2+1/3*x^3-1/4*x^4+\cdots = \sum_{n=1}^{\infty}\frac{(-1)^{n+1}x^n}{n}$$
What is this series S'(x)? (i.e., what function does it represent?) What happens at x=1?

5. Feb 26, 2012

### AlephZero

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

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. Feb 26, 2012

### vkash

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

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 syllabus(calculus is at lower portion) and give me a proper hint to approach this question.

Last edited: Feb 26, 2012
7. Feb 26, 2012

### pwsnafu

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

Hint: it's a Taylor series question.

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. Feb 26, 2012

### AlephZero

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

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.

9. Feb 26, 2012

### vkash

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

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.
is it know possible to solve it.

10. Feb 26, 2012

### vkash

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

what is this divergent and convergent series???

11. Feb 27, 2012

### D H

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

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. Feb 28, 2012

### checkitagain

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

No, $$e = \displaystyle\lim_{x\to 0}{(1 + x)}^{\frac{1}{x}}$$

$$e^x = 1 + \dfrac{x}{1!} + \dfrac{x^2}{2!} + \dfrac{x^3}{3!} + \ ...$$