Is it possible to sumate this series?

[Petar Mali]

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

$$C$$ - constant


Is it possible to sumate this series?

 

[quasar987]

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

 

[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?

 

[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.

 

[g_edgar]

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

 

[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,

[tex]\int_0^\infty dn~(2n+1)e^{-Cn(n+1)},[/itex]<br /> <br /> which is easily solved by substitution.<br /> <br /> Systematic corrections to the integral form can be computed using the Euler-Maclaurin formula:<br /> <br /> <a href="http://en.wikipedia.org/wiki/Euler-Maclaurin_formula" target="_blank" class="link link--external" rel="nofollow ugc noopener">http://en.wikipedia.org/wiki/Euler-Maclaurin_formula</a>[/tex]

 

[Petar Mali]

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,

[tex]\int_0^\infty dn~(2n+1)e^{-Cn(n+1)},[/itex]<br /> <br /> which is easily solved by substitution.<br /> <br /> Systematic corrections to the integral form can be computed using the Euler-Maclaurin formula:<br /> <br /> <a href="http://en.wikipedia.org/wiki/Euler-Maclaurin_formula" target="_blank" class="link link--external" rel="nofollow ugc noopener">http://en.wikipedia.org/wiki/Euler-Maclaurin_formula</a>[/tex]

[tex] <br /> Yes that is my problem. Sum is<br /> <br /> $$Q_1=\sum^{\infty}_{l=0}(2l+1)e^{-\frac{\theta_r}{\theta}l(l+1)$$<br /> <br /> where<br /> <br /> $$\theta_r=\frac{\hbar^2}{2I}$$<br /> <br /> Why we can say that if $$\theta>>\theta_r$$<br /> <br /> $$Q_1=\int^{\infty}_0dl(2l+1)e^{-\frac{\theta_r}{\theta}l(l+1)$$?[/tex]