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# (e-1)/(e+1)

## (e-1)/(e+1)

### Decimal expansion of (e-1)/(e+1)

The decimal expansion is

${\displaystyle \tanh({\tfrac {1}{2}})={\frac {e-1}{e+1}}=0.4621171572600097585023184836436725487302892803301130385527\ldots ,\,}$

where ${\displaystyle \scriptstyle \tanh(x)\,}$ is the hyperbolic tangent of ${\displaystyle \scriptstyle x\,}$. This gives the sequence of decimal digits (Cf. A160327)

{4, 6, 2, 1, 1, 7, 1, 5, 7, 2, 6, 0, 0, 0, 9, 7, 5, 8, 5, 0, 2, 3, 1, 8, 4, 8, 3, 6, 4, 3, 6, 7, 2, 5, 4, 8, 7, 3, 0, 2, 8, 9, 2, 8, 0, 3, 3, 0, 1, 1, 3, 0, 3, 8, 5, 5, 2, 7, 3, 1, 8, 1, 5, 8, 3, 8, 0, 8, 0, 9, 0, 6, 1, 4, 0, 4, 0, 9, 2, 7, 8, 7, ...}

### Continued fraction for (e-1)/(e+1)

The continued fraction is

${\displaystyle \tanh({\tfrac {1}{2}})={\frac {e-1}{e+1}}=0~+~{\cfrac {1}{2+{\cfrac {1}{6+{\cfrac {1}{10+{\cfrac {1}{14+{\cfrac {1}{18+{\cfrac {1}{22+{\cfrac {1}{26+{\cfrac {1}{\ddots }}}}}}}}}}}}}}}}\,}$

giving the sequence of partial quotients (Cf. A016825${\displaystyle \scriptstyle (n-1),\,n\,\geq \,1\,}$)

{0, 2, 6, 10, 14, 18, 22, 26, 30, 34, 38, 42, 46, 50, 54, 58, 62, 66, 70, 74, 78, 82, 86, 90, 94, 98, 102, 106, 110, 114, 118, 122, 126, 130, 134, 138, 142, 146, 150, 154, 158, 162, 166, 170, 174, 178, 182, ...}

## (e+1)/(e-1)

### Decimal expansion of (e+1)/(e-1)

The decimal expansion is

${\displaystyle \coth({\tfrac {1}{2}})={\frac {e+1}{e-1}}=2.1639534137386528487700040102180231170937386021507922725335\ldots ,\,}$

where ${\displaystyle \scriptstyle \coth(x)\,}$ is the hyperbolic cotangent of ${\displaystyle \scriptstyle x\,}$. This gives the sequence of decimal digits (Cf. A??????)

{2, 1, 6, 3, 9, 5, 3, 4, 1, 3, 7, 3, 8, 6, 5, 2, 8, 4, 8, 7, 7, 0, 0, 0, 4, 0, 1, 0, 2, 1, 8, 0, ...}

### Continued fraction for (e+1)/(e-1)

The continued fraction is

${\displaystyle \coth({\tfrac {1}{2}})={\frac {e+1}{e-1}}=2~+~{\cfrac {1}{6+{\cfrac {1}{10+{\cfrac {1}{14+{\cfrac {1}{18+{\cfrac {1}{22+{\cfrac {1}{26+{\cfrac {1}{30+{\cfrac {1}{\ddots }}}}}}}}}}}}}}}}\,}$

giving the sequence of partial quotients (Cf. A016825${\displaystyle \scriptstyle (n),\,n\,\geq \,0\,}$)

{2, 6, 10, 14, 18, 22, 26, 30, 34, 38, 42, 46, 50, 54, 58, 62, 66, 70, 74, 78, 82, 86, 90, 94, 98, 102, 106, 110, 114, 118, 122, 126, 130, 134, 138, 142, 146, 150, 154, 158, 162, 166, 170, 174, 178, 182, ...}

which happens to be the number of electrons per filled orbital of the atom.