[RESOLVED] Analytical expression (and proof) needed for limit of sum of series

[EvoRob]

Consider the following sum of a series:

s(n,X) = sum for i = 1 to n of 1/(n+nX+i-1)

Where, i and n are integers, and X is a free variable that may take any value >= 0.

For example, when n = 3, we have:
s(3,X) = 1/(3+3X) + 1/(3+3X+1) + 1/(3+3X+2)

I am trying to derive an analytical expression f(X) (not a summation) which gives the limit of the sum s(n,X) as n tends to infinity i.e. f(X) = s(infinity,X).

Suggestions as to an expression and proof would be most welcome.

Many thanks

Rob

[krtxmrtz]

Are you doing this for fun (the joy of math, I undertand ) or is it a school assignment?

[EvoRob]

Are you doing this for fun (the joy of math, I undertand ) or is it a school assignment?

I haven't been to school since the early 1980's

This series and the related simpler one in my earlier thread, are part of a task scheduling problem I'm working on. Any suggestions on how to find an analytical expression would be much appreciated.

Rob

[EvoRob]

I've managed to simplify the problem I'm trying to solve down to the following:

Let s(k) = sum for i = 1 to k of 1/((Wk/(1-W))+i-1) where W is the omega constant defined by the expression ln(1/W) = W (W = 0.567143), and i and k are integers

By numerical analysis, the limiting value of s(k) as k tends to infinity is W (omega)

Now I just need a proof that this is the case...

Thanks

Rob

[jemidiah]

This is definitely an interesting series, though I don't have enough experience with series to take more than shots in the dark.

It appears that as X gets large, f(x) ~= 1/x. Aside from that I don't see any immediately recognizable function or pattern that this would fit. I'm wondering, though, why do you need a closed-form expression for this? Brute force calculation of s(n, x) for large n doesn't take that long on a modern computer, if you only have to do a few calculations.

Edit: you beat me to posting. I'm still curious why you need a proof, though perhaps simply looking at the Maclaurin series for ln(1/x) or ln(1/(1+x)) will yield an something you can use here as in your other thread.

[EvoRob]

I'm still curious why you need a proof

I'm putting together a research paper, and an analytical proof and therefore an exact value would be nicer than just a numerical solution.

Thanks for your time and effort thinking about this. I'll have a look at the Maclaurin series as you suggest and see if I can make any headway using that.

Rob

[EvoRob]

After a lot of thought and a helpful hint from someone far better at maths than me, here is the solution:

s(n,X) = sum for i = 1 to n of 1/(n+nX+i-1)

Can be expressed as
s(n,X) = sum for i = 1 to n of 1/n. 1/(1+X+(i-1)/n)
Now this sum is recognisable as a left Riemann sum of the function y =1/x over the partition [1+X, 1+X+1] with n intervals of width 1/n starting at 1+X+(i-1)/n for i = 1 to n.

The limit of this sum as n tends to infinity is the integral of 1/x dx from 1+X to 2+X. Which is ln((2+X)/(1+X))

Rob

[krtxmrtz]

Nice and cute!

[jemidiah]

Very clever, cool