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

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

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

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

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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

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

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.

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I'm still curious why you need a proof

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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

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

Nice and cute!

Very clever, cool :)