I am trying to understand why $J_{-n}(x)=(-1)^n J_{n}(x)$.
In the following link http://oer.physics.manchester.ac.uk/PDEs/Notes/jsmath/Notesse37.html the author proves this by doing the following
(Edit: there is a typo in the link. $\Gamma(-n+k+1)!$ should be corrected to $\Gamma(-n+k+1)$, without the factorial sign. This correction was made in the expressions below.)
The definition of $J_n(x)$
$$ {J}_{n}(x) ={ \mathop{∑ }}_{k=0}^{∞} {{(−1)}^{k}\over k!(n + k)!}{\left ({x\over 2}\right )}^{n+2k}. $$
The proof: $J_{-n}(x)=(-1)^n J_{n}(x)$
$$ \begin{eqnarray}{ J}_{−n}(x)& =& {\mathop{∑ }}_{k=0}^{∞} {{(−1)}^{k}\over k!Γ(−n + k + 1)}{\left ({x\over 2}\right )}^{n+2k} %& \\ & =& {\mathop{∑ }}_{k=n}^{∞} {{(−1)}^{k}\over k!Γ(−n + k + 1)}{\left ({x\over 2}\right )}^{−n+2k}%& \\ & =& {\mathop{∑ }}_{l=0}^{∞}{{(−1)}^{l+n}\over (l + n)!l!}{\left ({x\over 2}\right )}^{n+2l} %& \\ & =& {(−1)}^{n}{J}_{ n}(x). %&(10.41) \\ \end{eqnarray} $$
There are two thing I don't understand about the proof. Could someone explain why
in the first line of the proof why is it $(\frac{x}{2})^{2k+n}$ not $(\frac{x}{2})^{2k-n}$.
Assuming the first line of the proof is right, why is it when the that we changed the lower limit of the sum from $k=0$ to $k=n$, $k! \Gamma(-n+k+1)$ remained unchanged.