# Is it possible to sumate this series?

1. Sep 3, 2009

### Petar Mali

$$\sum^{\infty}_{n=0}(2n+1)e^{-Cn(n+1)}$$

$$C$$ - constant

Is it possible to sumate this series?

2. Sep 3, 2009

### quasar987

If C is negative or 0, then no, obviously. If C>0, then yes...

3. Sep 4, 2009

### Petar Mali

Yes $$C>0$$. But I don't ask you if it convergate but can you summate analyticaly? Or you must use some numerical methods?

4. Sep 4, 2009

### quasar987

This is called finding a "closed form" for the series. I don't think one exists. So numerical approximation is the best one can do.

5. Sep 4, 2009

### g_edgar

A theta function or relative ... because of the quadratic in the exponent.

6. Sep 4, 2009

### Mute

That looks like the sum you need to do if you were to calculate the Partition function for a quantum mechanical dumbell. That sum cannot be computed in closed form. The typical approximations are to assume either low temperature ( $C \rightarrow \infty$), in which case the sum is approximately

$$1 + 3e^{-2C} + \ldots .$$

The other limit is high temperature, $C \rightarrow 0$, in which case the sum is approximately an integral,

$$\int_0^\infty dn~(2n+1)e^{-Cn(n+1)},[/itex] which is easily solved by substitution. Systematic corrections to the integral form can be computed using the Euler-Maclaurin formula: http://en.wikipedia.org/wiki/Euler-Maclaurin_formula 7. Sep 5, 2009 ### Petar Mali Yes that is my problem. Sum is [tex]Q_1=\sum^{\infty}_{l=0}(2l+1)e^{-\frac{\theta_r}{\theta}l(l+1)$$

where

$$\theta_r=\frac{\hbar^2}{2I}$$

Why we can say that if $$\theta>>\theta_r$$

$$Q_1=\int^{\infty}_0dl(2l+1)e^{-\frac{\theta_r}{\theta}l(l+1)$$?

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