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Is there an infinite series for the base-10 logarithm of 3 ($0.47712125471966243...$) that does not involve any irrational numbers? Bonus points if it involves the powers of 3, for a reason I will mention later.

We do have the following series for $\ln 3$: $\sum_{k=1}^\infty \frac{2^k}{k3^k}$. However, we would have to divide this by $\ln 10$, and although we have series for $\ln x$ in general, we would need to somehow divide the ln(3) series by the ln(10) series.

If it exists, my ultimate aim is to calculate the first digits of $^53 = 3^{3^{3^{3^3}}}$, as I sort of mentioned in my other recent post. This would require shifting the series in the same way as in Wikipedia's example of evaluating a spigot algorithm for ln(2), but with powers of 3 as the exponent of that number is a power of 3.

Allam A.
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If it exists, my ultimate aim is to calculate the first digits of. . .

I doubt that line of attack will work, personally. A naive analytic approach is most likely going to expand the number into terms that are still unmanageably large. You would be better off using modular arithmetic (and exploiting symmetries).

For expanding $\log_{10}x$ as a rational-coefficient power series, here is one method, which should work but would require quite tedious algebra if approached naively, so I would suggest using it in conjunction with a computer algebra system or a more sophisticated algebraic or combinatoric technique to streamline it. As $\log_{10}x=\ln x / \ln 10$, we can expand the natural logarithms individually as Taylor series and then consolidate the quotients/products as Cauchy products into a single power series. As the Taylor series of the natural logarithm about $x=1$ has radius of convergence $1$, we can use $\ln x=-\ln 1/x$ to connect $10$ to the radius of convergence.

Hence, $\log_{10}x=-\sum\frac1{k}{\left(1-x\right)^{k}}/\sum\frac1{j}{\left(\frac{9}{10}\right)^{j}}$ (within an appropriate radius of convergence), but we would prefer this as a product of power series instead of a quotient, to facilitate the algebra. To that end, we can use the geometric series $\frac1{1-S}=1+S+S^2+\ldots$ to expand the reciprocal as a power series. However, we need to extract some terms from the series so that $|S|<1$ in order to allow convergence. The first two terms, $\frac{9}{10}+\frac{1}{2}\left(\frac{9}{10}\right)^{2}=261/200$, suffice (barely) as they leave $0.997\ldots$ remaining in the series.

Therefore,

$$\log_{10}=\left(-\sum_{k\ge1}\frac{\left(1-x\right)^{k}}{k}\right)\cdot\left(\frac{200}{261}\sum_{j\ge0}\left(-\frac{200}{261}\right)^{j}S^{j}\right)$$

where $S=\sum_{i\ge3}\frac1{i}{\left(\frac{9}{10}\right)^{i}}$. This leaves us with a glut of unwieldy Cauchy products, the inner $S$ series in the $j$'th power and the product of the series over $k$ and $j$. And these can ultimately be combined into rational coefficients.

Jam
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  • Also, how can symmetries be used for this problem? I know the modular arithmetic trick (ab/c can be reduced to its fractional part by taking (ab mod c)/c), but it seems like it would still take trillions of steps to calculate the first digits of 3^^5 in this way. – Allam A. Dec 13 '23 at 16:40