IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

hsbhatt

sboosy: I thought that's a regular sum of geometric progression with c.r. = -r. and |r|<1

@apurva: I checked even with r² you get the same limit using S(1+r²). So there's no problem with the limit at all. Sorry it takes me some time to do these evaluations. 1 bcos its so long since i touched such stuff and 2. i have a job you know!

anchitsaini

neeraj_agarwal_1990

i think
2, -4, 6
is not a gp with common ratio -2

it is 2,-4,8
whose sum is 6 as calculated by u
hence the formula is valid for negative common ratio

anandghegde

Quite an interesting discussion!!!!!

elastiboysai

Lets call S as f(r)

clearly f(r)<0 for r>1 and f(r)>0 for r<1 (even no. of terms)

also f(r) depends on the number of terms,

for n tendin to infinity, the ans 1/4 is rt if its r<1

the summation formula holds good only if |r|<1

so the limit must be taken as r tends to 1 minus.

u r not justified in using it for r=1.

and importantly at r=1

, the function has value -3*n*(n+1)/2 as u suggested were n is d no of terms.

so it is finite n depds. on d no. of terms.

so i think its basically to do wid d discontinuity of the function

as discontinuous,

u are not justified in equating summation of some terms less dan 1 upto infin(wich turns out to be finite ) to anoder finite quantity!!

i.e

so fn is -ve for r>1 and even no.of terms.

fn is pos. for r>1 and odd no. of terms.

if u take r tends to 1 plus ull end up wid same limit as f(r) at r=1.

so LHL not = RHL for f(r).

wat u havdone will hold good in a case wen f(r) is continuous, doesn depend on no. of terms frm either side. (which im not sure is poss.)

i think das d prob here.

hsbhatt

elastiboysai: just let's say i am equating the LHL's of two equivalent expressions.

So we are still on the right side of mathematical law (pun unintended)

Also, S(n) = -n/2 when n is even and n/2+1 when n is odd.

apurviitjee2008

sorry Sir i had unnecesarily made some lenghty calculations and created confusion
Sir i too had this doubt on taking -r as the common ratio
i was not confident whether it was valid
and i still am not sure whether it is valid to take that
i had just taken it to be valid
and the point raised by elastiboy Sir has correctly pointed out that the sum of numbers again comes out to be negative
which is ODD

hsbhatt

I think its time to bring the curtains down on this discussion. I initiated it partly because it has a certain shock value and more because such discussions really take us back to the basics (like something sniper initiated)

The starting point is the series 1-2+3-4+5-... Many of you would have been puzzled because you know for sure that this series is divergent because for odd number of terms you get a positive sum and for even terms a negative number. And we get on the other hand, that the sum works out to a definite number 1/4. Usually, we tend to discard such a summation as being divergent and hence not worthy of our interest. But not some mathematicians.

Like some chap called Cesaro who insisted that 1-1+1-1+... had a sum. It equalled 1/2 !! He argued that if you take even terms you get zero, and if you take odd terms you get 1. So the sum averages out to 1/2.

A similar argument is used with the sum 1-2+4-8+... it "averages" out to 1/3.

Does this have any meaning? Yes, and in all places the real world. It appealed to the intutition of Leibnitz that when nature was confronted with such choices, she would choose their average. You have already heard that such a thing happens in quantum physics that the quantum wave function is actually a weighted average of possibilities.

This is just one method of summing up, one that is useful in such cases. There are other methods like Ramanujan Summation which was used in this particular case of 1-2+2-4 ..

And so, it is in physics, especially quantum physics where such divergent series routinely turn up and are dealt with in this way to produce results that agree with observations.

That's how mystical mathematics can get.

Please forgive me if I have wasted your time. But, if you want to read more on such stuff, read it on wikipedia. I found starting from keyword Abel summation gives a lot of info.

feynmann

Hey !! u are making a complete mess of it !!

There is no mumbo- jumbo in that qn .

U have performed an invalid step .

U evaluated the sum taking r

1 , which means that the sum is valid for r

1+ also , which is simply not the case .

The sum does not exist for r>1 , hence the limit does not exist !!

As simple as that !!

hsbhatt

Not quite so simple if you read my posts carefully. I say that the left limits of two equivalent expressions are being equated.

If you still want to argue on whether the limit is correct, just remember you are taking on Riemann, Abel, Cesaro, Liebnitz, Borel and not to speak of our own Ramanujan and a host of advanced calculus students. In short, it is a pretty well defined procedure. Maybe I omitted to say this, but the conditions under which such summations can be performed are well laid out which incidentally cover geometric series with r>1. No magic there!

And I suspect Feynmann himself would have used such a summation at the Shelter Island Conference.

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