Theorem: Series Expansion of Bessel Function of the First Kind

{{begin-eqn}}
{{eqn | l = \map {J_n} x
      | r = \dfrac {x^n} {2^n \, \map \Gamma {n + 1} } \paren {1 - \dfrac {x^2} {2 \paren {2 n + 2} } + \dfrac {x^4} {2 \times 4 \paren {2 n + 2} \paren {2 n + 4} } - \cdots}
      | c = 
}}
{{eqn | r = \sum_{k \mathop = 0}^\infty \dfrac {\paren {-1}^k} {k! \, \map \Gamma {n + k + 1} } \paren {\dfrac x 2}^{n + 2 k}
      | c = 
}}
{{end-eqn}}

Proof:
We employ Frobenius's method to find the solutions to the Bessel's Equation:
:$x^2 \dfrac {\d^2 y} {\d x^2} + x \dfrac {\d y} {\d x} + \paren {x^2 - n^2} y = 0$
for $n \ge 0$, in the form:
:$\ds \map y x = \sum_{k \mathop = 0}^\infty A_k x^{k + r}$
defined on $x > 0$, for some constants $r, A_i$, with $A_0 \ne 0$, which are to be determined.
Differentiating the expression {{WRT|Differentiation}} $x$:
{{begin-eqn}}
{{eqn | l = \map {y'} x
      | r = \sum_{k \mathop = 0}^\infty A_k \paren {k + r} x^{k + r - 1}
}}
{{eqn | l = \map {y' '} x
      | r = \sum_{k \mathop = 0}^\infty A_k \paren {k + r} \paren {k + r - 1} x^{k + r - 2}
}}
{{end-eqn}}
Substituting $y, y', y' '$ into Bessel's Equation:
{{begin-eqn}}
{{eqn | l = 0
      | r = x^2 \dfrac {\d^2 y} {\d x^2} + x \dfrac {\d y} {\d x} + \paren {x^2 - n^2} y
}}
{{eqn | r = x^2 \sum_{k \mathop = 0}^\infty A_k \paren {k + r} \paren {k + r - 1} x^{k + r - 2} + x \sum_{k \mathop = 0}^\infty A_k \paren {k + r} x^{k + r - 1} + \paren {x^2 - n^2} \sum_{k \mathop = 0}^\infty A_k x^{k + r}
}}
{{eqn | r = \sum_{k \mathop = 0}^\infty A_k \paren {k + r} \paren {k + r - 1} x^{k + r} + \sum_{k \mathop = 0}^\infty A_k \paren {k + r} x^{k + r} + \sum_{k \mathop = 0}^\infty A_k x^{k + r + 2} - n^2 \sum_{k \mathop = 0}^\infty A_k x^{k + r}
}}
{{eqn | r = \sum_{k \mathop = 0}^\infty A_k \paren {\paren {k + r} \paren {k + r - 1} + \paren {k + r} - n^2} x^{k + r} + \sum_{k \mathop = 0}^\infty A_k x^{k + r + 2}
}}
{{eqn | r = \sum_{k \mathop = 0}^\infty A_k \paren {\paren {k + r}^2 - n^2} x^{k + r} + \sum_{k \mathop = 0}^\infty A_k x^{k + r + 2}
}}
{{eqn | r = \sum_{k \mathop = 0}^\infty A_k \paren {\paren {k + r}^2 - n^2} x^{k + r} + \sum_{k \mathop = 2}^\infty A_{k - 2} x^{k + r}
      | c = Translation of Index Variable of Summation
}}
{{eqn | r = \sum_{k \mathop = 0}^\infty A_k \paren {\paren {k + r}^2 - n^2} x^k + \sum_{k \mathop = 2}^\infty A_{k - 2} x^k
      | c = $x^r \neq 0$
}}
{{end-eqn}}
Comparing the constant term on both sides:
{{begin-eqn}}
{{eqn | l = 0
      | r = A_0 \paren {r^2 - n^2}
}}
{{eqn | ll = \leadsto
      | l = r
      | r = \pm n
}}
{{end-eqn}}
Take $r = n$.
Comparing the rest of the coefficients:
{{begin-eqn}}
{{eqn | l = 0
      | r = A_1 \paren {\paren{n + 1}^2 - n^2}
}}
{{eqn | l = 0
      | r = A_1 \paren {2 n + 1}
}}
{{eqn | ll = \leadsto
      | l = A_1
      | r = 0
}}
{{eqn | l = 0
      | r = A_k \paren {\paren{n + k}^2 - n^2} + A_{k - 2}
      | c = for $k \ge 2$
}}
{{eqn | l = 0
      | r = A_k k \paren {2 n + k} + A_{k - 2}
}}
{{eqn | ll = \leadsto
      | l = A_k
      | r = - \dfrac 1 {k \paren {2 n + k} } A_{k - 2}
}}
{{end-eqn}}
From the recurrence relation above, we see that $A_k = 0$ for odd $k$, and:
{{begin-eqn}}
{{eqn | l = A_{2 k}
      | r = - \dfrac 1 {2 k \paren {2 n + 2 k} } A_{2 k - 2}
}}
{{eqn | r = \dfrac {-1} {2^2 k \paren {n + k} } A_{2 k - 2}
}}
{{eqn | r = \dfrac {\paren {-1}^2} {2^4 k \paren {k - 1} \paren {n + k}  \paren {n + k - 1} } A_{2 k - 4}
}}
{{eqn | r = \cdots
}}
{{eqn | r = \dfrac {\paren {-1}^k} {2^{2 k} k \paren {k - 1} \cdots \paren 1 \paren {n + k} \paren {n + k - 1} \cdots \paren {n + 1} } A_0
}}
{{eqn | r = \dfrac {\paren {-1}^k} {2^{2 k} k! \dfrac {\map \Gamma {n + k + 1} } {\map \Gamma {n + 1} } } A_0
}}
{{eqn | r = \dfrac {\paren {-1}^k} {2^{n + 2 k} k! \, \map \Gamma {n + k + 1} }
      | c = By choice of $A_0 = \dfrac {2^{- n} } {\map \Gamma {n + 1} }$
}}
{{end-eqn}}
Substituting this result to our original equation: 
{{begin-eqn}}
{{eqn | l = \map y x
      | r = \sum_{k \mathop = 0}^\infty A_k x^{k + n}
}}
{{eqn | r = \sum_{k \mathop = 0}^\infty A_{2 k} x^{2 k + n}
      | c = since all even terms vanish
}}
{{eqn | r = \sum_{k \mathop = 0}^\infty \dfrac {\paren {-1}^k} {2^{n + 2 k} k! \, \map \Gamma {n + k + 1} } x^{2 k + n}
}}
{{eqn | r = \sum_{k \mathop = 0}^\infty \dfrac {\paren {-1}^k} {k! \, \map \Gamma {n + k + 1} } \paren {\dfrac x 2}^{2 k + n}
}}
{{end-eqn}}
{{qed}}