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I am having difficulty computing the limit

$$\sum_{n=1}^\infty \exp(\beta n + \gamma \exp(\lambda n))$$

where $\beta<0$, $\gamma\in\mathbb R$ and $\lambda<0$. The constants $\beta, \gamma, \lambda$ are known. (I have to compute the above limit on a list of about $2^{17}$ values for $(\beta,\gamma,\lambda)$.) I think 2-4 decimal places of accuracy would be enough.

I can see that for $\gamma=0$ we have the limit of a geometric series, but when $\gamma\neq 0$ we have a decaying term inside the outer exponential.

After reading this post I picked representative values $\beta=-1/120, \gamma=1/10, \lambda=-1/12$ and tried computing the partial sums $S_N=\sum_{n=1}^N a_n$ where $a_n=\exp(\beta n + \gamma \exp(\lambda n))$. I also computed $\Delta(k)=S_{2^k}-S_{2^{k-1}}$, $Q(k)=\Delta(k)/\Delta(k-1)$ and the partial sums $\sum_{n=1}^N b_n$ for $b_n=\exp(\beta n)$. I get the following table, rounding to 4 decimal places:

\begin{array}{rrrrrrrr} \hline N=2^k & a_N & S_N & S_{2^{k-1}} & \Delta(k) & Q(k) & b_N & \sum_1^N b_n \\ \hline 2 & 1.0703 & 2.1576 & & & & 0.9835 & 1.9752 \\ 4 & 1.0391 & 4.2510 & 2.1576 & 2.0934 & & 0.9672 & 3.9177 \\ 8 & 0.9848 & 8.2685 & 4.2510 & 4.0175 & 1.9192 & 0.9355 & 7.7070 \\ 16 & 0.8985 & 15.7434 & 8.2685 & 7.4748 & 1.8606 & 0.8752 & 14.9169 \\ 32 & 0.7713 & 28.9846 & 15.7434 & 13.2412 & 1.7714 & 0.7659 & 27.9717 \\ 64 & 0.5869 & 50.4615 & 28.9846 & 21.4769 & 1.6220 & 0.5866 & 49.3961 \\ 128 & 0.3442 & 79.4425 & 50.4615 & 28.9810 & 1.3494 & 0.3442 & 78.3741 \\ 256 & 0.1184 & 106.4152 & 79.4425 & 26.9727 & 0.9307 & 0.1184 & 105.3468 \\ 512 & 0.0140 & 118.8927 & 106.4152 & 12.4775 & 0.4626 & 0.0140 & 117.8243 \\ 1024 & 0.0002 & 120.5456 & 118.8927 & 1.6529 & 0.1325 & 0.0002 & 119.4772 \\ 2048 & 0.0000 & 120.5691 & 120.5456 & 0.0235 & 0.0142 & 0.0000 & 119.5007 \\ 4096 & 0.0000 & 120.5691 & 120.5691 & 0.0000 & 0.0002 & 0.0000 & 119.5007 \\ 8192 & 0.0000 & 120.5691 & 120.5691 & 0.0000 & 0.0000 & 0.0000 & 119.5007 \\ \hline \end{array}

The last three values in the table for $a_N$ are 3.872941e-08, 1.499967e-15 and 2.249901e-30. This table leads me to conjecture that for this particular value of $(\beta, \gamma, \lambda)$, I find the limit to 2 decimal places for $N=2^{12}$.

So for each given value of $(\beta, \gamma, \lambda)$, I could compute $S_{2^k}$ for $k=13,14,\ldots$ until $\Delta(k)<10^{-4}$. Is this a reasonable algorithm for this particular problem?

I don't know much about extrapolation methods (e.g. Richardson). Are there particular features of the problem I could use to improve the speed/accuracy of the numerical method (substantially)?

For example, when $\gamma>0$ I see $b_n<a_n$ and the limit of the series in $b_n$ provides a lower bound for $S_\infty$. In the above table $a_n$ comes within 4 decimal places of $b_n$ for $n>=128$.

Community
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JS1204
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1 Answers1

5

Write $$\exp(\gamma \exp(\lambda n)) = \sum_{k=0}^\infty \dfrac{\gamma^k}{k!} \exp(k \lambda n)$$ Your sum becomes $$ \sum_{k=0}^\infty \dfrac{\gamma^k}{k!} \sum_{n=1}^\infty \exp((\beta + k \lambda) n) = \sum_{k=0}^\infty \dfrac{\gamma^k}{k!} \dfrac{\exp(\beta + k \lambda)}{1 - \exp(\beta + k \lambda)} $$ The sum on the right should converge quite quickly. For the values $\beta = -1/120$, $\gamma = 1/20$, $\lambda = -1/12$, you get an answer to within an absolute error of $.0001$ by going up to $k=2$.

Robert Israel
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