Divisors Sum Related Interesting Approximate Relation

Question

Working on Divisors Sum Efficient calulcation topic. Accidentaly discovered one interesting relation which is accurate up to $10^{17}$ order. $$\sum_{i=1}^{\infty}{\frac{\sigma(i)}{e^{i}}}\approx\frac{\pi^2}{6}-\frac{1}{2}+\frac{1}{24}$$

To get things more clear look at the below numbers:

$$\sum_{i=1}^{\infty}{\frac{\sigma(i)}{e^{i}}}=1.1866007335148928206...$$ $$\frac{\pi^2}{6}-\frac{1}{2}+\frac{1}{24}=1.1866007335148931031...$$

Just would like to share this nice relation, check if you know some paper about this and wondering if there are some similar known relations for other number theoretical functions:)

Just to clarify things this is not the only realation, but one of many, for example: $$\sum_{i=1}^{\infty}{\frac{\sigma(i)}{\sqrt{e^i}}}\approx\frac{2\pi^2}{3}-1+\frac{1}{24}$$

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EDITED

Accorging to @Greg Martin comment just realized that this is Lambert Series example.

Lets assume now $s>0$.

So the general rule is $$\sum_{i=1}^{\infty}{\frac{\sigma(i)}{e^{si}}}=\sum_{i=1}^{\infty}{\frac{i}{e^{si}-1}}$$

Using Euler-Maclaurin summation: $$\sum_{i=1}^{\infty}{\frac{i}{e^{si}-1}}=\int_{0}^{\infty}\frac{x}{e^{sx}-1}dx - \frac{1}{2}\big(\lim_{x\to0}\frac{x}{e^{sx}-1}+\lim_{x\to\infty}\frac{x}{e^{sx}-1}\big)+\frac{1}{12}\big(\lim_{x\to\infty}(\frac{x}{e^{sx}-1})'-\lim_{x\to0}(\frac{x}{e^{sx}-1})'\big)-\frac{1}{720}\big(\lim_{x\to\infty}(\frac{x}{e^{sx}-1})'''-\lim_{x\to0}(\frac{x}{e^{sx}-1})'''\big)+...$$

Here all the higher order derivatives are odd. If we look at the $\frac{x}{e^{sx}-1}$ all the odd derivatives $(\frac{x}{e^{sx}-1})^{(2k-1)}$ at $x\to0$ and $\infty$ are equal $0$ except the first derivative.

Here is the list of few odd derivatives:

$$\lim_{x\to0}\big(\frac{x}{e^{sx}-1}\big)=\frac{1}{s}; \lim_{x\to\infty}\big(\frac{x}{e^{sx}-1}\big)=0$$

$$\lim_{x\to0}\big(\frac{x}{e^{sx}-1}\big)'=-\frac{1}{2}; \lim_{x\to\infty}\big(\frac{x}{e^{sx}-1}\big)'=0$$

$$\lim_{x\to0}\big(\frac{x}{e^{sx}-1}\big)'''=0; \lim_{x\to\infty}\big(\frac{x}{e^{sx}-1}\big)'''=0$$

$$\lim_{x\to0}\big(\frac{x}{e^{sx}-1}\big)^{(5)}=0; \lim_{x\to\infty}\big(\frac{x}{e^{sx}-1}\big)^{(5)}=0$$

and integral equals to $$\int_{0}^{\infty}\frac{x}{e^{sx}-1}dx=\frac{\pi^2}{6s^2}$$

So finally we have:

$$\sum_{i=1}^{\infty}{\frac{\sigma(i)}{e^{si}}}=\frac{\pi^2}{6s^2}-\frac{1}{2s}+\frac{1}{24}$$

Conclusion Even after above formulas, the calculation shows that the real values are not matching. For example case for $s=1$. Any idea why this formula does not work?

Answer 1

Using, as you did, Euler-MacLaurin summation, the formula for infinite values of $n$ are a bit more complex.

For example, the expansion around $s=0$ leads to $$\sum_{i=1}^{\infty}{\frac{i}{e^{si}-1}}=\frac{\pi^2}{6s^2}-\frac{1}{2s}+\frac{1}{24}+$$ $$\frac{s^9}{632282112}\left(1-\frac{29332259 s^2}{62107500}+\frac{8569 s^4}{117000}-\frac{19774139 s^6}{2705040000}+O\left(s^8\right) \right)$$

Edit

After doing the Euler-MacLaurin expansion,assuming $s>0$ and $n \to \infty$, what is left is $$\color{blue}{\frac{\text{Li}_2\left(e^{-s}\right)}{s^2}-\frac{\log (1-e^{-s})}{s}+\frac{\sum _{s=0}^7 e^{k s}\,\, P_k(s)}{1209600 \left(e^s-1\right)^8}}$$ where the polynomials are of degree $7$. Starting from the constant term, the list of coefficients are

$$\left( \begin{array}{cc} k & \text{coefficients of } P_k(s) \\ 0 & \{-504000\} \\ 1 & \{3528000,100800,-5040,-1680,200,40,-7,-1\} \\ 2 & \{-10584000,-604800,20160,0,1600,960,-392,-120\} \\ 3 & \{17640000,1512000,-25200,15120,-3800,600,-1715,-1191\} \\ 4 & \{-17640000,-2016000,0,-26880,0,-3200,0,-2416\} \\ 5 & \{10584000,1512000,25200,15120,3800,600,1715,-1191\} \\ 6 & \{-3528000,-604800,-20160,0,-1600,960,392,-120\} \\ 7 & \{504000,100800,5040,-1680,-200,40,7,-1\} \\ \end{array} \right)$$

Computed for $s=1$, the decimal representation of the result is $$1.186600734454629573971129\cdots$$

The next level of Euler-MacLaurin summation formula would give $$\color{blue}{\frac{\text{Li}_2\left(e^{-s}\right)}{s^2}-\frac{\log (1-e^{-s})}{s}+\frac{\sum _{s=0}^9 e^{k s}\,\, P_k(s)}{239500800\left(e^s-1\right)^{10}}}$$ where the polynomials are of degree $9$.

Update

Working directly with $s=1$, which make calculations must faster, and playing with the successive orders of Euler-MacLaurin summation formula, the following results are $$\left( \begin{array}{cc} n & \text{result}_n \\ 1 & 1.186600734454629573971129 \\ 2 & 1.186600733577258935824311 \\ 3 & 1.186600733520094925774897 \\ 4 & 1.186600733515372855721243 \\ 5 & 1.186600733514926126548741 \\ 6 & 1.186600733514887173075080 \\ 7 & 1.186600733514887596144576 \\ 8 & 1.186600733514889998560738 \\ 9 & 1.186600733514891409585063 \\ 10 & 1.186600733514892131192378 \\ 11 & 1.186600733514892501128664 \\ 12 & 1.186600733514892697428227 \\ 13 & 1.186600733514892805224178 \\ 14 & 1.186600733514892865159871 \\ 15 & 1.186600733514892896968697 \\ \end{array} \right)$$

which seem to properly coincide with the given value of $\sum_{i=1}^{\infty}{\frac{\sigma(i)}{e^{i}}}$.