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A faster converging series

Before the new series just added, the two fastest converging series listed had ratios of successive terms which approached 1/9 and 1/16 in the limit. Thus their accuracy would increase by just under 1 decimal digit and just over 1.2 decimal digits respectively per term computed. For the new series, the limit of the ratio of successive terms is 1/961. Thus its accuracy would increase by just under 3 decimal digits per term computed. The new series may be derived as follows — first notice

${\displaystyle 2={\left({\frac {16}{15}}\right)}^{7}\cdot {\left({\frac {81}{80}}\right)}^{3}\cdot {\left({\frac {25}{24}}\right)}^{5}}$

which is easily verified by counting up the factors of 2, 3, and 5 (these are the only prime factors of these numbers). Thus

${\displaystyle \ln(2)=7\cdot \ln \left({\frac {16}{15}}\right)+3\cdot \ln \left({\frac {81}{80}}\right)+5\cdot \ln \left({\frac {25}{24}}\right)\,.}$

Then use the fact that

${\displaystyle \ln \left({\frac {n+1}{n}}\right)=\sum _{k=0}^{\infty }\,{\frac {2}{(2k+1)\,{(2n+1)}^{2k+1}}}}$

for n = 15, 80 and 24. JRSpriggs (talk) 00:44, 11 February 2014 (UTC)

This technique can be generalized to find the natural logarithm of many rational numbers. If A, B and C are integers, then

${\displaystyle \ln(2^{A}\cdot 3^{B}\cdot 5^{C})=(7A+11B+16C)\cdot \ln \left({\frac {16}{15}}\right)+(3A+5B+7C)\cdot \ln \left({\frac {81}{80}}\right)+(5A+8B+12C)\cdot \ln \left({\frac {25}{24}}\right)\,}$

which uses the same information about series as is required for calculating ln(2) this way. JRSpriggs (talk) 01:48, 12 February 2014 (UTC)

We could go a stage further and use

${\displaystyle 2=\left({\frac {1024}{1023}}\right)^{38}\cdot \left({\frac {243}{242}}\right)^{22}\cdot \left({\frac {125}{124}}\right)^{24}\cdot \left({\frac {121}{120}}\right)^{41}\cdot \left({\frac {961}{960}}\right)^{31}}$

which contain the prime factors: 2, 3, 5, 11, and 31. JRSpriggs (talk) 10:31, 21 April 2018 (UTC)

If we try to calculate an approximate value for the natural logarithm of two using just the first term in each part-series, we get

${\displaystyle \ln(2)\approx {\frac {76}{2047}}+{\frac {44}{485}}+{\frac {48}{249}}+{\frac {82}{241}}+{\frac {62}{1921}}=\mathbf {0.693\ 14} 4\ 056\ 986\ 778\ 6}$

with correct digits bolded. JRSpriggs (talk) 17:40, 26 April 2018 (UTC)

Justification of a BBP series formula

Here is a justification of the series formula

${\displaystyle \ln 2={\frac {2}{3}}+{\frac {1}{2}}\sum _{k=1}^{\infty }\left({\frac {1}{2k}}+{\frac {1}{4k+1}}+{\frac {1}{8k+4}}+{\frac {1}{16k+12}}\right){\frac {1}{16^{k}}}}$

which appears in the article.

1. From

${\displaystyle \ln {x}=\sum _{n=1}^{\infty }{1 \over {n}}\left({x-1 \over x}\right)^{n}}$

which appears in natural logarithm, we get

${\displaystyle \sum _{k=1}^{\infty }{\frac {1}{2k\,16^{k}}}={\frac {1}{2}}\sum _{k=1}^{\infty }{\frac {1}{k\,16^{k}}}={\frac {1}{2}}\ln \left({\frac {16}{15}}\right)\,.}$

2. The second term in the series can be converted to an integral for the time being:

${\displaystyle \sum _{k=1}^{\infty }{\frac {1}{(4k+1)\,16^{k}}}=2\sum _{k=1}^{\infty }{\frac {1}{4k+1}}\left({\frac {1}{2}}\right)^{4k+1}=2\int _{0}^{\frac {1}{2}}\sum _{k=1}^{\infty }x^{4k}\,\mathrm {d} x=2\int _{0}^{\frac {1}{2}}{\frac {x^{4}}{1-x^{4}}}\,\mathrm {d} x\,.}$

3. From

${\displaystyle \ln {x}=2\sum _{n=0}^{\infty }{1 \over {2n+1}}\left({x-1 \over x+1}\right)^{2n+1}}$

we get

${\displaystyle \sum _{k=1}^{\infty }{\frac {1}{(8k+4)\ 16^{k}}}=-{\frac {1}{4}}+\sum _{k=0}^{\infty }{\frac {1}{(2k+1)\ 4^{2k+1}}}={\frac {1}{2}}\ln \left({\frac {5}{3}}\right)-{\frac {1}{4}}\,.}$

4. The fourth term in the series becomes another integral

${\displaystyle \sum _{k=1}^{\infty }{\frac {1}{(16k+12)\,16^{k}}}=2\sum _{k=1}^{\infty }{\frac {1}{(4k+3)\,2^{4k+3}}}=2\int _{0}^{\frac {1}{2}}\sum _{k=1}^{\infty }x^{4k+2}\,\mathrm {d} x=2\int _{0}^{\frac {1}{2}}{\frac {x^{6}}{1-x^{4}}}\,\mathrm {d} x\,.}$

Adding the first and third parts gives

${\displaystyle {\frac {1}{2}}\ln \left({\frac {16}{15}}\right)+{\frac {1}{2}}\ln \left({\frac {5}{3}}\right)-{\frac {1}{4}}={\frac {1}{2}}\ln \left({\frac {16}{9}}\right)-{\frac {1}{4}}=\ln \left({\frac {4}{3}}\right)-{\frac {1}{4}}\,.}$

Adding the second and fourth parts gives

${\displaystyle 2\int _{0}^{\frac {1}{2}}{\frac {x^{4}}{1-x^{4}}}\,\mathrm {d} x+2\int _{0}^{\frac {1}{2}}{\frac {x^{6}}{1-x^{4}}}\,\mathrm {d} x=2\int _{0}^{\frac {1}{2}}{\frac {x^{4}+x^{6}}{1-x^{4}}}\,\mathrm {d} x=2\int _{0}^{\frac {1}{2}}{\frac {x^{4}(1+x^{2})}{(1-x^{2})(1+x^{2})}}\,\mathrm {d} x}$

${\displaystyle =2\int _{0}^{\frac {1}{2}}{\frac {x^{4}}{(1-x^{2})}}\,\mathrm {d} x=2\int _{0}^{\frac {1}{2}}{\frac {1}{(1-x^{2})}}\,\mathrm {d} x-2\int _{0}^{\frac {1}{2}}1+x^{2}\,\mathrm {d} x=\int _{0}^{\frac {1}{2}}{\frac {1}{(1-x)}}\,\mathrm {d} x+\int _{0}^{\frac {1}{2}}{\frac {1}{(1+x)}}\,\mathrm {d} x-2\left({\frac {1}{2}}+{\frac {1}{24}}\right)}$

${\displaystyle =-\ln {\frac {1}{2}}+\ln {\frac {3}{2}}-{\frac {13}{12}}=\ln 3-{\frac {13}{12}}\,.}$

Adding all four parts of the series together gives

${\displaystyle \ln \left({\frac {4}{3}}\right)-{\frac {1}{4}}+\ln 3-{\frac {13}{12}}=\ln 4-{\frac {4}{3}}\,.}$

Then the right hand side of the desired formula becomes

${\displaystyle {\frac {2}{3}}+{\frac {1}{2}}\left(\ln 4-{\frac {4}{3}}\right)=\ln 2\,}$

which is what was to be demonstrated. JRSpriggs (talk) 14:23, 13 February 2014 (UTC)

Why is this table here?

The lead has a table stuck onto the end, with no mention of it in the text, gving irrelevant information about the natural logs of various numbers, while this article is specifically about ln 2. Any objection to my removing the table? If not, I will move it to the article Natural logarithm. Loraof (talk) 02:31, 6 July 2017 (UTC)

Notice that the table is continued in the section Natural logarithm of 2#Bootstrapping other logarithms.

While this expands beyond the scope suggested by the article's title, I suspect that it would be better tolerated here than in the article natural logarithm which is more a general introduction and methods than about particular numerical values. JRSpriggs (talk) 05:15, 6 July 2017 (UTC)

How fast or slow is the convergence?

For the series

${\displaystyle \ln(2)={\frac {2}{3}}\sum _{k=0}^{\infty }{\frac {1}{9^{k}(2k+1)}},}$

I figure that one would have to carry the calculation out to the k=102 term in order to get 100 digits correct to the right of the decimal place.

For the series

${\displaystyle \ln(2)={\frac {14}{31}}\sum _{k=0}^{\infty }{\frac {1}{961^{k}(2k+1)}}\,+\,{\frac {6}{161}}\sum _{k=0}^{\infty }{\frac {1}{25921^{k}(2k+1)}}\,+\,{\frac {10}{49}}\sum _{k=0}^{\infty }{\frac {1}{2401^{k}(2k+1)}},}$

I figure that one would have to carry the calculation of the first part (for ln(16/15)) out to k=33, and the second part (for ln(81/80)) out to k=22, and the third part (for ln(25/24)) out to k=29. So while each part would have a significant speed up, taking them together would be only a marginal improvement. JRSpriggs (talk) 10:25, 15 April 2018 (UTC)

By one hundred digits correct, I mean that the error introduced by truncating the series after the k term would be less than ${\displaystyle \textstyle {\tfrac {1}{2}}10^{-100}}$. For the first series above, this means that

${\displaystyle {\frac {2}{3}}\sum _{m=k+1}^{\infty }{\frac {1}{9^{m}\cdot (2m+1)}}<{\frac {2}{3\cdot 9^{k}\cdot (2k+3)}}\sum _{j=1}^{\infty }{\frac {1}{9^{j}}}={\frac {2}{3\cdot 9^{k}\cdot (2k+3)\cdot 8}}<{\frac {1}{2}}10^{-100}}$.

${\displaystyle 10^{100}\cdot {\frac {2\cdot 2}{3\cdot (2k+3)\cdot 8}}<9^{k}}$.

${\displaystyle 100+\log _{10}{\frac {4}{4968}}<k\cdot \log _{10}9}$.

${\displaystyle {\frac {100-3.0941215958}{0.9542425094}}=101.5526739<k}$.

That is, k = 102.
In the other series, I allowed each of the three parts to have one third of the total permitted error. JRSpriggs (talk) 06:16, 17 April 2018 (UTC)

None of the series currently listed at Natural logarithm of 2#Rising alternate factorial would converge quickly enough to allow one to calculate one hundred digits correctly within the foreseeable future even with the most powerful computers. That is, as a practical matter, they do not converge. The first one listed would require one to go to the googolth term and the last would require about ten to the seventeenth power terms. Similarly, for series list at Natural logarithm of 2#Other series representations. I do not understand the Riemann-zeta function, so I cannot comment on that section. The series at Natural logarithm of 2#Binary rising constant factorial would converge, but painfully slowly, requiring hundreds of terms. JRSpriggs (talk) 04:06, 18 April 2018 (UTC)

Applying the three general series to the square of the square-root of 2

See Natural logarithm#Series for the justification of the three general series.

1. Applying the Taylor series gives:

${\displaystyle \ln(2)=\ln(({\sqrt {2}})^{2})=2\ln({\sqrt {2}})=2\sum _{n=1}^{\infty }{\frac {(-1)^{n-1}({\sqrt {2}}-1)^{n}}{n}}.}$

Then I use the fact that

${\displaystyle {\sqrt {2}}-1={\frac {1}{{\sqrt {2}}+1}}}$

to get

${\displaystyle \ln(2)=2\sum _{n=1}^{\infty }{\frac {(-1)^{n-1}}{({\sqrt {2}}+1)^{n}n}}.}$

This is preferable to the first form because it avoids cancellation which would reduce the number of correct significant digits. See loss of significance.

2. Applying the second general series gives:

${\displaystyle \ln(2)=2\sum _{n=1}^{\infty }\left({\frac {{\sqrt {2}}-1}{\sqrt {2}}}\right)^{n}{\frac {1}{n}}.}$

Using

${\displaystyle {\frac {{\sqrt {2}}-1}{\sqrt {2}}}={\frac {1}{({\sqrt {2}}+1){\sqrt {2}}}}={\frac {1}{2+{\sqrt {2}}}}}$

gives

${\displaystyle \ln(2)=2\sum _{n=1}^{\infty }{\frac {1}{(2+{\sqrt {2}})^{n}n}}.}$

3. Applying the fast (third) general series gives:

${\displaystyle \ln(2)=2\sum _{k=0}^{\infty }\left({\frac {{\sqrt {2}}-1}{{\sqrt {2}}+1}}\right)^{2k+1}{\frac {2}{2k+1}}.}$

Using

${\displaystyle {\frac {{\sqrt {2}}-1}{{\sqrt {2}}+1}}={\frac {1}{({\sqrt {2}}+1)^{2}}}={\frac {1}{3+2{\sqrt {2}}}}}$

and

${\displaystyle (3+2{\sqrt {2}})^{2}=17+12{\sqrt {2}}}$

gives

${\displaystyle \ln(2)={\frac {4}{3+2{\sqrt {2}}}}\sum _{k=0}^{\infty }{\frac {1}{(17+12{\sqrt {2}})^{k}(2k+1)}}.}$

OK? JRSpriggs (talk) 05:27, 21 April 2018 (UTC)

Continued Fraction

I'm not sure how the notability criteria are applied to content such as this, but would it not be reasonable to include the simple continued fraction expansion of this number on the page? That's certainly what I just came here looking for, and I was surprised that it's missing, when such content appears in, for example: Pi#Continued fractions. The entries of the simple continued fraction of ln(2) are certainly documented in the expected place: OEIS A016730. -GTBacchus 17:00, 5 May 2018 (UTC)

There are two continued fraction in the section Natural logarithm of 2#Other representations. And you could copy in and specialize the continued fractions in Natural logarithm#Continued fractions. You are free to segregate these into a new "Continued fractions" section. And to add any for which you can provide reliable secondary sources. Continued fractions are not my thing, so I am not going to do it. JRSpriggs (talk) 04:15, 6 May 2018 (UTC)

Thank you; I've made an edit. :) -GTBacchus 20:15, 13 June 2018 (UTC)

Alternating Harmonic Series

I'd have said that the most interesting fact (especially for math students) about ln(2) is that it is the sum of a named series, the "alternating harmonic series". I think it should be right in the intro in its expanded form, rather than just being grouped with the other series representations. --Farry (talk) 09:14, 29 June 2018 (UTC)

${\displaystyle \ln(2)=1-{\frac {1}{2}}+{\frac {1}{3}}-{\frac {1}{4}}+{\frac {1}{5}}-\cdots }$

Off-topic

JRSpriggs starts an edit war over several counts at once without an attempt to discuss. There are two principal dubious items: the digression about decimal logarithms and the table of ln n, n ∊ ℕ. “lack of a better place for it” is hypocrisy – firstly, there are such other places as Wikibooks and, secondly, not an sufficient argument anyway to keep off-topic in an article.

Moreover, JRSpriggs restores a mediocre schoolchild’s whitespace in “ln (2)” which is already not far from vandalism – check the changes before saving the edit, next time. Incnis Mrsi (talk) 07:48, 20 July 2019 (UTC)

Please avoid the personal attacks.

I have edited and used this article for more than five years. This appears to be your first edit of it. You removed material which has been useful to me in calculations which I have carried out.

I am not aware of any other article in wikipedia which contains or reasonably could contain this table. Perhaps this article should be renamed to "Table of natural logarithms" or "List of natural logarithms", but no one has previously suggested that.

By reverting your edit, I was not saying that every single change you made was unacceptable, just that the edit as a whole was unacceptable. Personally, I like to have a blank between the "ln" and the "(2)", but I would not have reverted you if that was the only change you made.

Also, I would not object if you moved the table down to the section Natural logarithm of 2#Bootstrapping other logarithms where the table is continued for some prime numbers. JRSpriggs (talk) 10:12, 21 July 2019 (UTC)

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