Exponential Sum Approximation into a Hyperbolic Cotangent I'm working on a problem relating to the length and elasticity of a polymer chain(Kardar, SM/Particles Ch. 2 Pr. 9). When trying to check my answer against a solutions manual, it gives the following approximation for $N\gg1$:
$$\langle R^2\rangle =\sum^N_{m,n} a^2e^{-|m-n|/\xi}\approx a^2N\Big[1+2\frac{e^{-1/\xi}}{1-e^{-1/\xi}}\Big]=a^2N\coth{\frac{1}{2\xi}}$$
It mentions that the sum decays exponentially around each point and that end effects are asymptotically negligible for $N\rightarrow\infty$, which both make sense to me. Nevertheless, I'm still not how the approximation works mathematically.
Any help would be appreciated.
 A: Let's look at $\sum^N_{m,n} e^{-|m-n|}$ (taking $\zeta=1$ for now for simplicity) and try to match series:

*

*For m = 1, we have $(1+e^{-1}+e^{-2}+\cdots+e^{-(N-1)})$.

*For m = 2, we have $e^{-1}+(1+e^{-1}+e^{-2}+\cdots+e^{-(N-1)})-e^{-(N-1)}.$

*For m = 3, we have $e^{-2}+e^{-1}+(1+e^{-1}+e^{-2}+\cdots+e^{-(N-1)})-e^{-(N-1)}-e^{-(N-2)}.$

*$\cdots$

*For m = N, we have $(e^{-(N-1)}+e^{-(N-2)}+\cdots+e^{-1})+(1+e^{-1}+e^{-2}+\cdots+e^{-(N-1)})-(e^{-(N-1)}+e^{-(N-2)}+\cdots+e^{-1}).$
Now we add those all together.
First, consider that we have $N$ copies of $(1+e^{-1}+e^{-2}+\cdots)$, or $$\sum^N_{r=0}(e^{-1})^r=N\frac{1-e^{-(N+1)}}{1-e^{-1}}=N\frac{e-e^{-N}}{e-1}.$$
We could assume $e^{-N}$ is negligible, giving a term $\frac{Ne}{e-1}$.
Second, from the terms preceding the series that appears $N$ times, we have the series $(N-1)e^{-1}+(N-2)e^{-2}+\cdots+e^{-(N-1)}=\sum_{r=1}^{N-1}re^{-(N-r)}$, or $$e^{-N}\sum_{r=1}^{N-1}re^r=e^{-N}\left[e\frac{1-Ne^{N-1}+(N-1)e^{N}}{(1-e)^2}\right]=\frac{e^{-N+1}-N+(N-1)e}{(e-1)^2}.$$
If we assume $e^{-N+1}$ to be negligible and take $N\approx(N-1)$, the second term reduces to $\frac{N}{e-1}.$
Third, from the terms following the series that appears $N$ times, we have the series $(N-1)e^{-(N-1)}+(N-2)e^{-(N-2)}+\cdots+e^{-1}=\sum_{r=1}^{N-1}r(e^{-1})^r$, or
$$\sum^{N-1}_{r=1}r(e^{-1})^r=\left(\frac{1}{e}\right)\frac{1-Ne^{-(N-1)}+(N-1)e^{-N}}{(1-e^{-1})^2}.$$
Assuming $e^{-N}$ and $e^{-(N-1)}$ to be negligible, the third term is $\frac{e}{(e-1)^2}$, which is negligible compared to the first and second terms of $O(N)$.
By adding $\frac{Ne}{e-1}$ and $\frac{N}{e-1}$, we obtain the desired term, which is $$N\left(1+2\frac{e^{-1}}{1-e^{-1}}\right)=N\left(1+\frac{2}{e-1}\right);$$
the case of arbitrary $\zeta$ is just a matter of replacing each $e$ term with  $e^{1/\zeta}$. Does this make sense?
