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If the sequence is $\frac{1}{1+e}, \frac{1}{1+e^2}, \frac{1}{1+e^3}, \cdots, \frac{1}{1+e^n}$, how can I find the sum or product of the above sequence? That is $$\sum_{k=1}^{n} \frac{1}{1+e^k}$$ and $$\prod_{k=1}^{n} \frac{1}{1+e^k}$$ The sequence is not an arithmetic, geometric, or harmonic progression, so what can I do?

julio_es_sui_glace
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Dev Rathore
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  • Try entire series developpement? For the second one you should apply $\ln$ to it – julio_es_sui_glace Oct 18 '23 at 17:06
  • Sorry for interfering. Pls show the full solution. It's a very interesting question – Shivam Kumar Oct 18 '23 at 17:12
  • I don't think there is any easy formula, if you only need an estimate take the integral. Wolfram alpha says i is a digamma function. – trula Oct 18 '23 at 17:18
  • pls send me the full solution. – Dev Rathore Oct 18 '23 at 17:21
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    I couldn't find a duplicate question, but I found some potentially relevant questions: 1 (answers first question, if it were an infinite series) and 2 (is related to the second question, and has no answer, but does link to another related post, and a relevant paper). – Theo Bendit Oct 18 '23 at 17:35

1 Answers1

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In terms of the q-digamma function, there is an explicit solution for the sum $$S_n=\sum_{k=1}^{n} \frac{1}{e^k+1}=\psi _{\frac{1}{e}}^{(0)}(1-i \pi )-\psi _{\frac{1}{e}}^{(0)}(n-i \pi +1)$$

This is nice but not very pactical if one does not access this special function.

The continuous case is much more simple $$T_n=\int_{1}^{n} \frac{dk}{e^k+1}=n-1-\log \left(\frac{e+1}{e^n+1}\right)$$

The limit

$$L=\underset{n\to \infty }{\text{limit }}\frac{S_n}{T_n}=\frac{\log \left(\frac{e-1}{e}\right)+\psi _{\frac{1}{e}}^{(0)}(1-i \pi )}{\log \left(\frac{e+1}{e}\right)}=1.48171172619\cdots$$ is very quickly atteinned from above (for example, if $n=10$, the ratio is $1.48184$).

So, an approximation $$S_n \sim L\, T_n$$

For $n=2^p$, some numerical results $$\left( \begin{array}{ccc} p & \text{approximation} & \text{exact result} \\ 3 & 0.4636665402055017829693530 & 0.4639683019375577689499164 \\ 4 & 0.4641633490165810934671019 & 0.4641634502684112933990670 \\ 5 & 0.4641635157612409366699013 & 0.4641635157612523310631363 \\ 6 & 0.4641635157612597013124978 & 0.4641635157612597013124978 \\ \end{array} \right)$$

An empirical correction could be $$S_n \sim L\, T_n+e^{-(n+\frac 1 {10})}$$ which would give $$\left( \begin{array}{ccc} p & \text{approximation} & \text{exact result} \\ 3 & 0.4639700793435806496302185 & 0.4639683019375577689499164 \\ 4 & 0.4641634508426180246671108 & 0.4641634502684112933990670 \\ 5 & 0.4641635157612523956807584 & 0.4641635157612523310631363 \\ 6 & 0.4641635157612597013124978 & 0.4641635157612597013124978 \\ \end{array} \right)$$

Similarly, using the q-Pochhammer symbol, the product

$$P_n=\prod_{k=1}^{n} \frac{1}{e^k+1}=\frac{2}{(-1;e)_{n+1}}$$ could be compared to $$Q_n=\exp\left(\text{Li}_2\left(-e^n\right)+\text{Li}_2\left(-\frac{1}{e}\right) +\frac{3+\pi ^2}{6} \right)$$

They are very highly correlated but all further analysis would be totally emperical.

Claude Leibovici
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