# Infinite Sum Question (q-harmonic?)

1. Mar 20, 2008

### re8

I have been working with the following series $\sum_{n=0}^\infty \frac{x^n}{a+x^n}$, where $0<x < 1$. I had a feeling it might be related to q-harmonic series, but I really have no idea:-)

I am looking for either a good analytic approximation, or even some idea of the sensitivity of this sum to the parameters x and a. But I would grateful if anyone could point me in the right direction.

2. Mar 20, 2008

### CRGreathouse

Interesting. The function seems to be well-behaved enough, at least for 'reasonable' values of x and a. In what context did you find this? I wasn't able to find anything in the 'normal places', but I might have missed something.

What kinds of values do you have for a and x?

3. Mar 20, 2008

### re8

Context

x can be pretty much anything between 0 and 1; I need to check a to see what kinds of restrictions I have on it, but I think it is also between 0 and 1.

In case you are curious, I'm an economist and it is from a model we formulated:-) It essentially represents the future benefit from varying a certain action x in the model, and we want to show that this future benefit is dominated by the cost today of the action (hence the action is not worthwhile) - hence the need for some idea of the impact of changing x on this sum. I can give you more detail if you like. A bound on the derivative of the sum w.r.t. x would be a start, although these changes aren't necessarily infinitesimal. I tried obvious things like using the initial terms, but it wasn't quite enough.

I have simulated it for a very fine grid of values for x between 0 and 1 and I get the results I want, but I would like to see if I can get an analytic result.

4. Mar 21, 2008

### CRGreathouse

I don't know of an analytic result. For a fixed value of either x or a it probably wouldn't be hard, but for both... nothing comes to mind.

How close to the extremes can the values come? Would x = 0.9999 be common? As long x isn't too close to 1 and a isn't too close to 0 it's easy to calculate this.

5. Mar 21, 2008

### etaoin shrdlu

I haven't checked this, so I have probably gotten some of the regions of validity wrong, but here are my thoughts on approximating the sum.

Write $N = \frac{\log a}{\log x}$ so that your series becomes $\sum_{n=0}^\infty \frac{x^n}{a+x^n} = \sum_{n=0}^\infty \frac{1}{1+x^{N-n}}$. Over all integer $n$, this curve is a sigmoid, and $N$ is the knee of the curve (value $1/2$) where terms transition from "near 1" to "near 0"; you can approximate or bound the series separately in these two regimes.

Things are especially convenient if you can restrict $2N$ to an integer; $\frac{1}{1+x^{N-n}}+\frac{1}{1+x^{N-(2N-n)}}=1$, so terms $n$ and $2N-n$ of the series now sum to $1$ and the sum from $0$ to $2N$ is $\frac{2N+1}{2}$ exactly, and you only have to consider the tail of the sum. But even if $2N$ is not an integer, this is a pretty good approximation for the first $\round{2N}$ terms of the sum (including a fraction of the final term).

If $1/a$ or $1/x$ is large, then in the tail of the sum ($n>2N$) you can reasonably approximate $\frac{1}{1+x^{N-n}} \approx x^{n-N}-x^{2(n-N)}+\cdots$ using the first few terms of a geometric series; now just sum these to get an estimate of the tail of the series. This estimate isn't very good when $a$ and $x$ are near $1$; but for $x$ near $1$ the terms vary slowly with $n$, and you can approximate the sum by an integral (which evaluates to a hypergeometric function).

6. Mar 22, 2008

### re8

Thanks - I'll try this

Thanks - I'll try that