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# Sqrt(e)

The square root of
 e
has the Maclaurin series expansion
${\displaystyle {\sqrt {e}}=\sum _{n=0}^{\infty }{\frac {1}{(2n)!!}}=\sum _{n=0}^{\infty }{\frac {1}{2^{n}n!}},\,}$
where
 n!!
is the double factorial.
 2√  e
is transcendental since
 e
is transcendental.

## Value

The decimal expansion of
 2√  e
is (A019774)
${\displaystyle {\sqrt {e}}=1.6487212707001281468486507878141635716537761007101480115750\ldots }$

with continued fraction

${\displaystyle {\sqrt {e}}=1~+~{\cfrac {1}{{\textbf {1}}+{\cfrac {1}{1+{\cfrac {1}{1+{\cfrac {1}{{\textbf {5}}+{\cfrac {1}{1+{\cfrac {1}{1+{\cfrac {1}{{\textbf {9}}+{\cfrac {1}{1+{\cfrac {1}{\ddots }}}}}}}}}}}}}}}}}},\quad {\frac {1}{{\sqrt {e}}-1}}={\textbf {1}}~+~{\cfrac {1}{1+{\cfrac {1}{1+{\cfrac {1}{{\textbf {5}}+{\cfrac {1}{1+{\cfrac {1}{1+{\cfrac {1}{{\textbf {9}}+{\cfrac {1}{1+{\cfrac {1}{\ddots }}}}}}}}}}}}}}}}\,}$
giving the sequence (A058281
 (n), n   ≥   0,
for
 2√  e
; A058281
 (n), n   ≥   1,
for 1/(sqrt(e)-1)) with pattern
 a (3k + 1) = 4k + 1, a(n) = 1
otherwise
{1, 1, 1, 1, 5, 1, 1, 9, 1, 1, 13, 1, 1, 17, 1, 1, 21, 1, 1, 25, 1, 1, 29, 1, 1, 33, 1, 1, 37, 1, 1, 41, 1, 1, 45, 1, 1, 49, 1, 1, 53, 1, 1, 57, 1, 1, 61, 1, 1, 65, 1, 1, 69, 1, 1, 73, 1, 1, 77, ...}

## 1/(sqrt(e)-1)

The decimal expansion of 1/(sqrt(e)-1) is (A113011)

${\displaystyle {\frac {1}{{\sqrt {e}}-1}}=1.54149408253679828413110344447251463834045923684188210947413695663\ldots }$
${\displaystyle {\frac {1}{{\sqrt {e}}-1}}=1+{\cfrac {2}{3+{\cfrac {4}{5+{\cfrac {6}{7+{\cfrac {8}{\ddots }}}}}}}}}$