For positive n, the ordinary differential equation
$$y''+\dfrac{1}{x}y'+\left(1-\dfrac{n^2}{x^2}\right)y=0$$
has as a solution the Bessel function of order n, $J_n\left(x\right)=x^n\sum^{\infty}_{k=0}a_kx^k$
Determine $a_1$ by substituing the seris for the Bessel function into the differential equation.
Im unsure of how to determine $a_1$. I have so far $$\sum^{\infty}_{k=2}\left[a_k\left[\left(k+n\right)\left(k+n-1\right)+\left(k+n\right)-n^2\right]+a_{k-2}\right]x^{k+n-2} +\left[a_o\left(n\left(n-1\right)+n-n^2\right)\right]x^{n-2}+\left[a_1\left(n\left(1+n\right)+\left(1+n\right)-n^2\right)\right]x^{n-1}=0$$
For $x^{n-1}$:
$a_1\left(2n+1\right)=0$ so $\Rightarrow n=-\dfrac{1}{2}$
For $x^{k+n-2}$:
$a_k\left[\left(k+n\right)\left(k+n\right)-n^2\right]=-a_{k-2}$
So we have the reccurence relation
$a_k=\dfrac{-a_{k-2}}{k\left(k-1\right)}$
They are some hitchs :
First, do not forget the condition $n>1$ : If not, some terms doesn't exist in your previous equations.
Second, $n$ is an integer and cannot be equal to $\frac{1}{2}$. So the conclusion is simple : $$a_1=0\qquad \text{in case of } n>1$$
Note : The cases $n=0$ and $n=1$ are easy to to treat directly ( No need for using the full series : the first terms are sufficient for a simpler writting) . I don't give all details. You will find this :
Case $n=0$ : $ J_0(x)=1-\frac{x^2}{4}+...$ Thus $a_1=0$ (also obvious because $ J_0(x)$ is an even function).
Case $n=1$ : $J_1(x)=\frac{x}{2}+...$ Thus $a_1=\frac{1}{2}$