I encountered a proof question as the following shows:
I can solve the first sub-question by using divisional integration method for $\Gamma(x+1)$.
But I got problems on the second sub-question. Even though I could write down and somewhat simplify the $J_{m-1}(x)$ and $J_{m+1}(x)$, if I am doing right, they are:
$$J_{m-1}(x)=\sum_{n=0}^\infty\frac{(-1)^n}{n!\Gamma(m+n)}\left(\frac x2\right)^{m-1+2n}\;\;\;\;(1)$$
$$J_{m+1}(x)=\sum_{n=0}^\infty\frac{(-1)^n}{n!\Gamma(m+n+2)}\left(\frac x2\right)^{m+1+2n}\;\;\;\;(2)$$
I can also get the derivation of $J_{m}(x)$:
$$2\frac{J_m(x)}{dx}=\sum_{n=0}^\infty\frac{(-1)^n}{n!\Gamma(m+n+1)}\left(\frac x2\right)^{m-1+2n}(m+2n)\;\;\;\;(3)$$
But I got confused just from the start. As you can see, although in (1) and (3) the power of x is the same, in (2) and (3) they have a $\left(\frac x2\right)^2$ difference. As x is a variable, how can $(1)-(2)=(3)$? So I totally have no ideas.
I do not know whether my thoughts are somewhere wrong or maybe the question did not clarify clearly somewhere. But could you give me some hints or comments on this problem? Thank you very much!
Just write $$J_{m+1}(x)=\sum_{n=0}^\infty\frac{(-1)^n}{n!\Gamma(m+n+2)}\left(\frac x2\right)^{m+1+2n} =\sum_{n=1}^\infty\frac{(-1)^{n-1}}{(n-1)!\Gamma(m+n+1)}\left(\frac x2\right)^{m-1+2n} =\sum_{n=1}^\infty\frac{n(-1)^{n-1}}{n!\Gamma(m+n+1)}\left(\frac x2\right)^{m-1+2n} =\sum_{n=0}^\infty\frac{n(-1)^{n-1}}{n!\Gamma(m+n+1)}\left(\frac x2\right)^{m-1+2n}$$ and notice that $$\frac1{\Gamma(m+n)}+\frac n{\Gamma(m+n+1)}=\frac{m+2n}{\Gamma(m+n+1)}.$$