My strategy was to express

¥ å n=0

nr xn

using the power series for 1/(1-x)^p for different integers p; it turns out that p must range from 1 to r+1. If you let A_r,p be the coefficient of 1/(1-x)^p, some interesting patterns emerge. I have found that A_r,r+1=r!, A_r,r=-(r+1)!/2, and A_r,r-1=(3r+2)(r+1)!/24. I have succeeded in proving these. However, more confusingly, it seems that

Ar,p=

p-1 å s=0

p-1Cs (-1)r+s(s+1)r

. This formula is very weird. For example, in the special case p=r+1 it should be equal to r!; I have spent two weeks trying to prove this and failed. Can anybody help?

I will send a table of values for A_r,p for r < 10, if anybody is interested.

David Loeffler

Okay... here is a sketchy reply.

Yes, the series you mention does sum to 6. This is proved in exactly the way you mention.

Consider (1-x)^-1 = 1 + x + x² + ...

What happens if we differentiate and then multiply by x? (in otherwords what happens when we apply the operator D = x d/dx). If you do this you should be able to see how to get the types of series you would like and by looking at what the operator does you should be able to get relations between the A(r,p).

Second method... What about if we just differentiate a few times? Well we will get a series with general term of the form n*(n+1)*...*(n+c). So now we have to work out how to express things like n^r in terms of the above types of functions (the coefficients here will almost match the ones you want). Well there is a good quick method for doing things like this - it is called Newton's method . What you do is you write the series you want to get in a line. Then underneath write the differences between terms. Underneath this write the differences and so on:

0 1 4 9 16 25 ...
1 3 5 7 9 ...
2 2 2 2 ...
0 0 0 ...

Now the numbers along the left hand diagonal [well, it would be diagonal if spaces came out on this system!!] (0,1,2,0,0,...) give the coefficients you want (almost). Try it! I'm not going to prove why this works or even what I mean by almost... if you fiddle around you will soon get it!

Now do this with the first row being

0^r 1^r 2^r ...

and you will get sums like your A(r,p) sum you write down. So this gives a proof.

I won't prove why the special case gives r! because it will just amount to reproving everything above. In other words, if someone had asked me to prove the A(r,r+1) identity totally independent of me seeing where it comes from then I would have effectively reconstructed the fact that it is a coefficient in the expansions you use.

Let me know how you get on... and I will elaborate on anything that you don't get from this reply. I just don't want to spoil your enjoyment by doing too much of it in this reply!!

AlexB.