So the $n$-th Bessel function $J_n(x)$ expressed as a power series is
$$J_n(x) = (\frac{x}{2})^n \sum_{k=0}^{\infty}\frac{(-1)^k}{k! (n+k)!} (\frac{x}{2})^{2k}$$
This function is supposed to solve the following ODE:
$$x^2y'' + xy' + (x^2 - n^2)y = 0$$
So I have tried to differentiate this power series and substitute it back into the ODE to see if it really equals $0$ (I am not interested in solving the ODE using Frobenius Method to arrive at the power series)
But I have been trying to do this for literally hours now and I can't seem to get anywhere.
Here is the best with what I have come up with so far:
$$J_n'(x)= \frac{1}{2} n (\frac{x}{2})^{n-1} \cdot \frac{J_n(x)}{(\frac{x}{2})^{n}} + (\frac{x}{2})^n \sum_{k=1}^{\infty}\frac{(-1)^k \cdot k}{k! (n+k)!} (\frac{x}{2})^{2k-1} \\= \frac{1}{2} n (\frac{x}{2})^{n-1} \cdot \frac{J_n(x)}{(\frac{x}{2})^{n}} - (\frac{x}{2})^{n+1} \sum_{k=0}^{\infty}\frac{(-1)^k }{k! (n+k + 1)!} (\frac{x}{2})^{2k} \\= \frac{1}{2} n (\frac{x}{2})^{n-1} \cdot \frac{J_n(x)}{(\frac{x}{2})^{n}} - J_{n+1}(x)$$
$$\Rightarrow x \cdot J_n'(x) = n \cdot J_n(x) - x\cdot J_{n+1}(x)$$
Then differentiating this implicit equation gives:
$$x \cdot J_n''(x) + J_n'(x) = n \cdot J_n'(x) - x \cdot J_{n+1}'(x) - J_{n+1}(x)$$
$$\Rightarrow x^2 \cdot J_n''(x) + x \cdot J_n'(x) = nx \cdot J_n'(x) - x^2 \cdot J_{n+1}'(x) - x \cdot J_{n+1}(x)$$
So then
$$nx \cdot J_n'(x) - x^2 \cdot J_{n+1}'(x) - x \cdot J_{n+1}(x) = n^2 J_n(x) - x^2 J_n(x)$$
And I just can't seem to find why the last equation holds.
Did I do something wrong? If not can anyone please give me a hint?
$$\begin{align} & {{J}_{n}}\left( x \right)={{\left( \frac{x}{2} \right)}^{n}}\sum\nolimits_{k=0}^{\infty }{\frac{{{\left( -1 \right)}^{k}}}{k!\left( n+k \right)!}{{\left( \frac{x}{2} \right)}^{2k}}}=\sum\nolimits_{k=0}^{\infty }{\frac{{{\left( -1 \right)}^{k}}}{k!\left( n+k \right)!}{{\left( \frac{x}{2} \right)}^{n+2k}}}=\sum\nolimits_{k=0}^{\infty }{\frac{{{\left( -1 \right)}^{k}}}{{{2}^{n+2k}}k!\left( n+k \right)!}{{x}^{n+2k}}} \\ & Now\,set\ {{a}_{k}}=\frac{{{\left( -1 \right)}^{k}}}{{{2}^{n+2k}}k!\left( n+k \right)!}\,,So\ {{J}_{n}}\left( x \right)=\sum\nolimits_{k=0}^{\infty }{{{a}_{k}}{{x}^{n+2k}}}\, \\ & \text{substituting}\,{{J}_{n}}\left( x \right)\text{ in the differential equation yields:} \\ & {{x}^{2}}{{\left( \sum\nolimits_{k=0}^{\infty }{{{a}_{k}}{{x}^{n+2k}}} \right)}^{\prime \prime }}+x{{\left( \sum\nolimits_{k=0}^{\infty }{{{a}_{k}}{{x}^{n+2k}}} \right)}^{\prime }}+\left( {{x}^{2}}-{{n}^{2}} \right)\left( \sum\nolimits_{k=0}^{\infty }{{{a}_{k}}{{x}^{n+2k}}} \right) \\ & {{x}^{2}}\left( \sum\nolimits_{k=0}^{\infty }{{{a}_{k}}\left( n+2k \right)\left( n+2k-1 \right){{x}^{n+2k-2}}} \right)+x\left( \sum\nolimits_{k=0}^{\infty }{{{a}_{k}}\left( n+2k \right){{x}^{n+2k-1}}} \right)+\left( {{x}^{2}}-{{n}^{2}} \right)\left( \sum\nolimits_{k=0}^{\infty }{{{a}_{k}}{{x}^{n+2k}}} \right) \\ & \sum\nolimits_{k=0}^{\infty }{{{a}_{k}}\left( n+2k \right)\left( n+2k-1 \right){{x}^{n+2k}}}+\sum\nolimits_{k=0}^{\infty }{{{a}_{k}}\left( n+2k \right){{x}^{n+2k}}}+\sum\nolimits_{k=0}^{\infty }{{{a}_{k}}{{x}^{n+2k+2}}}-\sum\nolimits_{k=0}^{\infty }{{{n}^{2}}{{a}_{k}}{{x}^{n+2k}}} \\ & \sum\nolimits_{k=0}^{\infty }{{{a}_{k}}\left( \left( n+2k \right)\left( n+2k-1 \right)+\left( n+2k \right)-{{n}^{2}} \right){{x}^{n+2k}}}+\sum\nolimits_{k=0}^{\infty }{{{a}_{k}}{{x}^{n+2k+2}}} \\ & \sum\nolimits_{k=0}^{\infty }{4{{a}_{k}}k\left( n+1 \right){{x}^{n+2k}}}+\sum\nolimits_{k=0}^{\infty }{{{a}_{k}}{{x}^{n+2k+2}}} \\ & \sum\nolimits_{k=1}^{\infty }{4{{a}_{k}}k\left( n+1 \right){{x}^{n+2k}}}+\sum\nolimits_{k=0}^{\infty }{{{a}_{k}}{{x}^{n+2k+2}}} \\ & \sum\nolimits_{k=0}^{\infty }{4{{a}_{k+1}}\left( k+1 \right)\left( n+1 \right){{x}^{n+2\left( k+1 \right)}}}+\sum\nolimits_{k=0}^{\infty }{{{a}_{k}}{{x}^{n+2k+2}}} \\ & \sum\nolimits_{k=0}^{\infty }{\left( 4{{a}_{k+1}}\left( k+1 \right)\left( n+1 \right)+{{a}_{k}} \right){{x}^{n+2k+2}}}=0\ \quad because\ 4{{a}_{k+1}}\left( k+1 \right)\left( n+1 \right)+{{a}_{k}}=0\ for\,all\ k. \\ \end{align} $$