# Sum of a series

1. Sep 17, 2004

### Parth Dave

If you have a series:
1^2 - 3^2 + 5^2 - 7^2 +... (-1)^(n-1)(2n-1)^2

how can you find a formula to represent the sum of the series. I have absolute no idea how to approach this question. The textbook doesn't show any examples of question where the found an equation for the sum of a series. They have a question however that is similar to this, that is

for the series:
1^2 + 3^2 + 5^2 + 7^2 + 9^2+...(2n-1)^2, the sum is:
(n/3)(2n-1)(2n+1)

I don't know how or if that helps. The textbook actually says to "guess" a summation formula and than prove it. So, any help would be appreciated.

2. Sep 17, 2004

### mathwonk

here's a trick that ahs been suggested elsewhere: assume the formula is cubic, like the one given in the other problem. i.e. just assume it looks like
an^3 + bn^2 + cn + d = f(n), and solve for a,b,c,d.

i.e. just plug in n=1 and f(1) = 1, and n=2 and f(2) = -8, n= 3 and f(3) = 17, and n=4 and f(4) = -32. see what you get.

3. Sep 17, 2004

### Parth Dave

Hmm.. i tried regressing a cubic equation on my calculator, using those 4 data points. It did give me a graph that was perfect, however only over that interval. It didn't work for anything greater than n = 4.

4. Sep 17, 2004

### Parth Dave

I was thinking that if you want to graph it as a polynomial, than the only way that it could work is if your polynomial was to the nth degree since it keeps oscillating between positive and negative.

5. Sep 17, 2004

### Gokul43201

Staff Emeritus
Hmmm...it's strange that a cubic didn't work !?! But your argument about n inflections makes sense...I'm sure I'm not thinking straight.

Here's another way to do this :

Call the required sum S(n). You know the sum for the other series; call that S'(n). Now add the two series :

S(n) + S'(n) = 2(1^2 + 5^2 + 9^2 + 13^2 + ...) the other terms cancel off.

Let's call this new sum S"(n) = 2*SUM[t"(n)^2], where t"(n) = 4n-3. Or t"(n)^2 = 16n^2 - 24n + 9

So, S"(n)/2 = 16*SUM[n^2] - 24*SUM[n] + 9*SUM[1] = 16n(n+1)(2n+1)/6 - 24n(n+1)/2 + 9n

Now simply find S"(n) - S'(n) = S(n), and you are home.

Last edited: Sep 17, 2004
6. Sep 18, 2004

### metacristi

If the general term of a series ( here (-1)^(n-1)(2n-1)^2;n ≥ 1 ) formed a linear sequence then it would be possible to use the theory of linear sequences to find the sum of the series (as I've shown here:https://www.physicsforums.com/showthread.php?p=300719#post300719).

I do not know whether it form a linear sequence in this case,at first sight it doesn't seem easy to find such a linear recurrence.I think it's much easier to split the sum (how it was already suggested) and take into account whether n is even or odd.

Write the series as:

If:

1. n=even then:

S[even]=1+{∑from k=1 to {(n/2)-1} (4k+1)2} - {∑from k=1 to n/2 (4k-1)2}

2.n=odd;

S[odd]=1+{∑from k=1 to (n-1)/2 (4k+1)2} - {∑from k=1 to (n-1)/2 (4k-1)2}

Finally after some calculations (tedious indeed,but it's plain algebra,anyway easy if one know the sum of 12+22+...+n2 and 1+2+...+n) --->

[edit to add]actually-by also rewriting in an appropriate form the sum for S[even]-it is not even needed to know the sum of 12+22+...+n2 [/edit]!

S[n=even]=-2n2 (1)

S[n=odd]=1+2[n2-1] (2)

For example if:

n=3 ---> S=1-9+25=17;using (2) ---> S=1+2*[9-1]=17 the same.

n=4 --->S=1-9+25-49=-32;by using (1) ---> S=-2*16=-32 the same.Everything seems OK (I hope,I haven't checked the results for more cases).

Last edited: Sep 19, 2004
7. Sep 18, 2004

### mathwonk

well i just solved the linear equations i suggested above by hand, and it seems to work, at least if you solve them correctly, which took me a while, after a few missteps inputting the wrong numbers. i.e. if the last term is (-1)^n (2n+1)^2,

then the sum seems to be given by the cubic 1 -62n + 71 n^2 -18 n^3.

I was encouraged by the fact that the more astute solutions offered by others also proved that the formula had degree <= 3.

if you want mkore ugly details, you get f(0) = 1, f(1) = -8, f(2) =17, and f(3) = -32. so the equations are:

a = 1, a+b+c+d = -8, a+2b+4c+8d = 17, a +3b +9c+27d = -32.

then you just solve them, e.g.by putting the coefficient matrix in echelon form. (After all these years of teaching that stuff, however, i still did it hook or crook by hand. I see now though that would make it look easier.)

Last edited: Sep 18, 2004
8. Sep 18, 2004

### Parth Dave

Mathwonk, that only seems to work for n = [0,3]. Everything outside of that doesn't work.

Gokul, I have absolutely no idea why your method doesn't work. But when I did what you said I couldn't get it to work. Hopefully I'm not just making some algebraic mistakes, I'll double check that to make sure.

Metacristi, as far as I can tell your method does work. It worked for the first n = [0,10] and I'm gonna prove it works for any value of n.

Thx for all the help guys, damn this question took long but i guess in the end it was worth it.

9. Sep 18, 2004

### mathwonk

i assuemd gokul's method worked which implied there was a solution of degree 3. i suspect his breaks down where he firt uses ..., instead of actually writing out the last term.

now metachristi's method does work i guess if you say so, but does not solve the problem of giving a single formula for all cases both even and odd does it? if it does, what is the formula, and what is its degree in n?

10. Sep 18, 2004

### Parth Dave

I have no idea if you can write it as a single formula. However, I don't see why it is necessary to do so.

11. Sep 18, 2004

### mathwonk

well, that's what you asked for, that's all I meant. i.e. you did not ask for the sum of the series, you asked for "a formula" that gives the sum.

12. Sep 19, 2004

### metacristi

In the general case an inifinite alternative series is convergent only if the limit for n-> ∞ from the closed forms of the sum (for n=odd and n=even) are finite and equal.For example the series a-a+a-a+.....+(-a)n-1+.... diverges for it has the sum '0' for n=even and 'a' for n=odd.That is we should anyway calculate the sums for the two cases and it's clear that the formulas in 'n' are different.Thus I do not think it is a good approach trying to solve these alternative series by using the theory of linear sequences before a convenient splitting of the sum (for example one having only the positive terms the other the negative ones) for the simple reason that it might not exist a single formula (I say 'might' for I do not have a formal proof for that).

Last edited: Sep 19, 2004
13. Sep 19, 2004

### mathwonk

well, after several tries, i still do not see what convergence has to do with this problem.

however the reason there is no polynomial formula for the sum seems to be that the partial sums of the series oscillates infinitely often between positive and negative values, as no polyonmial ever does.

i.e. there is no polynomial in n, that is positive for all odd values of n and negative for all even values of n, or vice versa.

so the problem, as posed, has no solution as a finite polynomial formula.

Last edited: Sep 19, 2004
14. Sep 19, 2004

### mathwonk

so just multiply metacristi's formula that works for even n, by the factor absolute value of cos(npi/2) and multiply his formula for the odd case, by absolute value of cos([n+1]pi/2).

that should give one "formula" that works for all n.

i.e. |cos(npi/2)|(-2n^2) + |cos([n+1]pi/2)| (1+2[n^2-1]).

if you do not like absolute values you can multiply by (1/2)[1 plus or minus cos(npi)].

15. Sep 19, 2004

### Gokul43201

Staff Emeritus
Haven't followed all that's been happening here, but it looks like I've screwed up somewhere. Ignore my posts. Will check back later...when I have more time.

Sorry, if I've caused immense (or even moderate) grief.

16. Sep 19, 2004

### mathwonk

not really i was just passing the blame for my error. my apologies.

I am a little puzzled that the texbook would ask this question,at least not in that form, unless they meant us to see there can be no polynomial solution.

or unless they had discussed this point before, or given examples where more than one formula was used for different cases.

on the other hand this is a very interesting problem, and once it is established this is the level of challenge of this textbook, one should put ones open - ended - thinking cap on when attacking their problems in future.

Last edited: Sep 19, 2004
17. Sep 19, 2004

### Parth Dave

The textbook gives absolutely no examples of how to generate a formula for the sum of the series. This was actually in the proof by induction section. I think they wanted you to look at patterns from previous proofs to help you come up with it. However, i couldn't see any patterns (nor could any of my classmates).

Now that i have one solid formula.. i embark on another journey: one to prove it true using mathematical induction.. let the fun begin.

And dont worry gokul, no grief caused. You jsut opened my mind to a different method of problem solving .

18. Sep 19, 2004

### mathwonk

suggestion. prove both of metacristi's formulas by induction. then mine is automatic.

19. Sep 20, 2004

### metacristi

How do you prove [formally] that 'the series oscillates infinitely often between positive and negative values, as no polyonmial ever does.'?It seems to me that you could prove that by...computing the closed forms for the partial sums and their limits for n tending to infinity,as I've indicated above.But this is not at all valid for all alternative series.I brought about the discussion about convergence (though basically I should have presented only that particular example) exactly because it offers a reason (not counting as a formal proof but a strong reason notwithstanding) to not attempt to find a single formula...Frankly speaking I do not understand what's your point,you are betting on a dead horse here.

Last edited: Sep 20, 2004
20. Sep 23, 2004