Continued Fraction for $\log 2$

Summary: Applying the §369 reciprocal-series template to the alternating harmonic series $\log 2=1-1/2+1/3-1/4+\cdots$ produces a continued fraction whose partial numerators are the squares $1,1,4,9,16,25,\dots$ and whose partial denominators are all $1$ — a structurally simple companion to [[brouncker-formula|Brouncker’s continued fraction for $4/\pi$]].

Sources: chapter18 (§369 Example I).

Last updated: 2026-05-11

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The identity

The alternating-harmonic series $\log 2=1-1/2+1/3-1/4+1/5-\cdots$ has the form $1/A-1/B+1/C-1/D+\cdots$ with $A=1,B=2,C=3,D=4,\dots$. By Template II of §369 — set $b=A=1$, $c=B-A=1$, $d=C-B=1$, $e=D-C=1$, $\dots$ — the partial numerators become $\alpha =1,\beta =A^2=1,\gamma =B^2=4,\delta =C^2=9,ϵ=D^2=16,\dots$ and all partial denominators are $1$.

Therefore

$\boxed{\log 2=\frac{1}{1+\frac{1}{1+\frac{4}{1+\frac{9}{1+\frac{16}{1+\frac{25}{1+\cdots }}}}}}}$

with partial numerators $1,1,2^2,3^2,4^2,5^2,\dots$ (i.e., the squares of $1,1,2,3,4,5,\dots$) and partial denominators all $1$.

Convergence

Like the Brouncker formula, this CF inherits slow ($\sim 1/n$) convergence from its parent series — the partial numerators $k^2$ grow at exactly the rate that cancels the geometric decay that bounded partial numerators would have produced. Faster convergence for $\log 2$ comes from the §120 fast variant $\log \left(\frac{1+x}{1-x}\right)=2(x+x^3/3+x^5/5+\cdots )$ evaluated at $x=1/3$ rather than from this CF.

Related pages

• continued-fraction-series-equivalence — the §369 reciprocal-series template Euler applies here
• logarithmic-series — source of the alternating-harmonic series for $\log 2$
• brouncker-formula — sister identity, partial numerators odd squares
• chapter-18-on-continued-fractions