EDIT:
I think the OP was referring to the recurrence relation $$\frac{d}{dx} \left(x^{-n}j_{n}(x) \right)=-x^{-n}j_{n+1}(x). \tag{1} $$
From $(1)$ it follows immediately that $$\begin{align}\int_{0}^{\infty}x^{-n} j_{n+1}(x) \, dx &= -x^{-n} j_{n}(x) \Bigg|^{\infty}_{0} \\ &=-x^{n} \sqrt{\frac{\pi}{2x}} \, J_{n+\frac{1}{2}}(x) \Bigg|^{\infty}_{0} \\ &= \lim_{x \downarrow 0} \, x^{-n} \sqrt{\frac{\pi}{2x}} \, J_{n+\frac{1}{2}}(x) \tag{2} \\ &=\lim_{x \downarrow 0} \, x^{-n} \sqrt{\frac{\pi}{2x}}\left(\frac{1}{\Gamma\left(n + \frac{3}{2}\right)} \left(\frac{x}{2}\right)^{n+ \frac{1}{2}} +\mathcal{O}\left(x^{n+ \frac{5}{2}} \right)\right) \tag{3} \\ &= \sqrt{\pi} \, 2^{-n-1} \, \frac{1}{\Gamma \left(n+ \frac{3}{2}\right)} \\ &= \sqrt{\pi} \, 2^{-n-1} \, \frac{2^{n+1}}{\sqrt{\pi} (2n+1)!!} \tag{4}\\ &= \frac{1}{(2n+1)!!}. \end{align}$$
$(1)$: http://dlmf.nist.gov/10.51#E3
$(2)$: https://en.wikipedia.org/wiki/Bessel_function#Asymptotic_forms
$(3)$: https://en.wikipedia.org/wiki/Bessel_function#Bessel_functions_of_the_first_kind:_J.CE.B1
$(4)$: http://mathworld.wolfram.com/DoubleFactorial.html (2)
This might not be the particular approach you were seeking, but it can be evaluated using a property of the Mellin transform.
Ramanujan's master theorem states that if $f(x)$ has an expansion of the form $$ f(x) = \sum_{k=0}^{\infty} \frac{\phi(k)}{k!} (-x)^{k} ,$$
then
$$ \int_{0}^{\infty} x^{s-1} f(x) \, dx = \Gamma(s) \phi(-s) $$
for the values of $s$ for which the integral converges.
The hypergeometric representation of the Bessel function of the first kind of order $\alpha$ is $$ J_{\alpha}(x) = \frac{(\frac{x}{2})^{\alpha}}{\Gamma(\alpha +1)} \ {}_0F_{1} \left(\alpha +1; - \frac{x^{2}}{4} \right) = \frac{(\frac{x}{2})^{\alpha}}{\Gamma(\alpha +1)} \sum_{k=0}^{\infty} \frac{\Gamma(\alpha + 1)}{\Gamma(\alpha +1+k)} \frac{(-\frac{x^{2}}{4})^{k}}{k!} .$$
So for $s+n+1 >0$ and $s <2$, we have
$$ \begin{align} \int_{0}^{\infty} x^{s-1} j_{n+1}(x) \, dx &= \int_{0}^{\infty} x^{s-1} \sqrt{\frac{\pi}{2x}} \, J_{n+\frac{3}{2}} (x) \, dx \\ &= \int_{0}^{\infty} x^{s-1} \sqrt{\frac{\pi}{2x}} \frac{(\frac{x}{2})^{n + \frac{3}{2}}}{\Gamma(n +\frac{5}{2})} \, {}_0F_{1} \left(n + \frac{5}{2}; - \frac{x^{2}}{4} \right) \, dx \\ &= \frac{\sqrt{\pi}}{2^{n+2}} \frac{1}{\Gamma(n+\frac{5}{2})} \int_{0}^{\infty} x^{s+n} \, {}_0F_{1} \left(n + \frac{5}{2}; - \frac{x^{2}}{4} \right) \, dx \\ &= \frac{\sqrt{\pi}}{2^{n+2}} \frac{1}{\Gamma(n+\frac{5}{2})} \int_{0}^{\infty} (2 \sqrt{u})^{s+n} \, {}_0F_{1} \left(n + \frac{5}{2}; - u \right) \, \frac{du}{\sqrt{u}} \\ &= \frac{ \sqrt{\pi} \ 2^{s-2}}{\Gamma(n+\frac{5}{2})} \int_{0}^{\infty} u^{\frac{s+n+1}{2}-1} \, {}_0F_{1} \left(n + \frac{5}{2}; - u \right) \, du \\ &= \frac{ \sqrt{\pi} \ 2^{s-2}}{\Gamma(n+\frac{5}{2})} \, \Gamma \left(\frac{s+n+1}{2} \right) \frac{\Gamma(n+ \frac{5}{2})}{\Gamma(n + \frac{5}{2} - \frac{s+n+1}{2})} \\ &= \sqrt{\pi} \ 2^{s-2} \, \Gamma \left(\frac{s+n+1}{2} \right) \frac{1}{\Gamma(\frac{4-s+n}{2})} . \end{align}$$
Now let $s= -n+1$.
Then
$$ \begin{align} \int_{0}^{\infty} x^{-n} j_{n+1}(x) \ dx &= \sqrt{\pi} \ 2^{-n-1} \Gamma (1) \, \frac{1}{\Gamma \left(n + \frac{3}{2} \right)} \\ &= \sqrt{\pi} \ 2^{-n-1} \frac{2^{n+1}}{\sqrt{\pi} (2n+1)!!} \tag{1} \\ &= \frac{1}{(2n+1)!!} \end{align}$$
$(1)$: http://mathworld.wolfram.com/DoubleFactorial.html (2)
I was interested by the transcendental equation. Letting $x=ka$, for stability reasons, I prefered to change its writng and consider that we look for the zero of function
$$f(x)=\left(1-x^2\right) \sin (x)-x \cos (x)$$ Since it is odd, discarding the trivial solution $x=0$, focus on the positive roots which are closer and closer to multiples of $\pi$.
Written as series around $x=k \pi$,
$$f(x)=-\pi k-\pi ^2 k^2 t-\frac{3\pi k}{2} t^2+\frac{\pi ^2 k^2-4}{6}
t^3+\frac{7\pi k}{24} t^4+\frac{16-\pi ^2 k^2}{120}
t^5-\frac{11\pi k}{720} t^6+O\left(t^{7}\right)$$ where $t=x-k \pi$. Now, using series reversion
$$x_{(k)}=\pi k-\frac{1}{\pi k}-\frac{5}{3 \pi ^3 k^3}-\frac{73}{15 \pi ^5 k^5}-\frac{1826}{105 \pi
^7 k^7}+O\left(\frac{1}{k^9}\right)$$ which seems to work quite well.
$$\left(
\begin{array}{ccc}
k & \text{estimation} & \text{solution} \\
1 & 2.747869217 & 2.743707270 \\
2 & 6.116769339 & 6.116764264 \\
3 & 9.316615752 & 9.316615629 \\
4 & 12.48593738 & 12.48593737 \\
5 & 15.64386611 & 15.64386611
\end{array}
\right)$$
Edit
Considering that we look for the first non trivial zero of function
$$f(x)=j_n(x)-\frac{x }{n}\,j_{n-1}(x)$$ we have
$$f'(x)=\sqrt{\frac{\pi }{2}}\,\,\,\frac{x^2-n(n+1) }{n \,x^{3/2}}J_{n+\frac{1}{2}}(x)$$
$$f''(x)=\sqrt{\frac{\pi }{2}}\,\,\,\frac{(n+1) \left(x^2-n(n-1) \right)J_{n+\frac{1}{2}}(x)-x \left(x^2-n (n+1)\right)J_{n+\frac{3}{2}}(x) }{n \,x^{5/2} }$$
The first derivative cancels at
$$x_*=\sqrt{n(n+1)}$$ To get an estimate, expand as series to get
$$x_0=x_*+\sqrt{-2\frac {f(x_*)}{f''(x_*)}}$$ that is to say
$$x_0=\sqrt{n(n+1)} +\sqrt{\sqrt{n(n+1)}\,\,\frac{ j_{n-1}\left(\sqrt{n(n+1)}
\right)}{j_n\left(\sqrt{n(n+1)} \right)}-n}$$ Iterations of Newton method
$$x_{k+1}=x_k+\frac{x_k \left(x_k J_{n-\frac{1}{2}}(x_k)-n
J_{n+\frac{1}{2}}(x_k)\right)}{\left(x_k^2-n(n+1)\right) J_{n+\frac{1}{2}}(x_k)}$$
A few values for comparison
$$\left(
\begin{array}{cccc}
n & x_0 & x_1 & x_2 & \text{solution} \\
1 & 2.6691616 & 2.7445999 & 2.7437074 & 2.7437074 \\
2 & 3.8798178 & 3.8702541 & 3.8702386 & 3.8702386 \\
3 & 5.0331291 & 4.9740292 & 4.9734204 & 4.9734204 \\
4 & 6.1579411 & 6.0635123 & 6.0619498 & 6.0619498 \\
5 & 7.2649512 & 7.1428326 & 7.1402287 & 7.1402288 \\
6 & 8.3595285 & 8.2144886 & 8.2108446 & 8.2108444 \\
7 & 9.4448169 & 9.2801189 & 9.2754678 & 9.2754680 \\
8 & 10.522842 & 10.340863 & 10.335248 & 10.335242 \\
9 & 11.594999 & 11.397549 & 11.391017 & 11.391017 \\
10 & 12.662292 & 12.450800 & 12.443395 & 12.443395 \\
20 & 23.176232 & 22.866090 & 22.851803 & 22.851767 \\
30 & 33.541865 & 33.167335 & 33.148172 & 33.148111 \\
40 & 43.834852 & 43.410759 & 43.387766 & 43.387766 \\
50 & 54.083230 & 53.618117 & 53.591940 & 53.592598 \\
60 & 64.300908 & 63.800426 & 63.771501 & 63.771384 \\
70 & 74.495928 & 73.964130 & 73.932777 & 73.932645 \\
80 & 84.673409 & 84.113369 & 84.079832 & 84.079686 \\
90 & 94.836838 & 94.250979 & 94.215451 & 94.215292 \\
100 & 104.98871 & 104.37900 & 104.34163 & 104.34031 \\
200 & 206.13568 & 205.34898 & 205.29815 & 205.29790 \\
300 & 306.94240 & 306.03338 & 305.97340 & 305.97309 \\
400 & 407.58540 & 406.57967 & 406.51253 & 406.51216 \\
500 & 508.12878 & 507.04171 & 506.96858 & 506.96818 \\
600 & 608.60398 & 607.44601 & 607.36768 & 607.36725 \\
700 & 709.02905 & 707.80782 & 707.72487 & 707.72440 \\
800 & 809.41545 & 808.13681 & 808.04967 & 807.51428 \\
900 & 909.77091 & 908.43955 & 908.34858 & 907.81456 \\
1000 & 1010.1010 & 1008.7207 & 1008.6262 & 1008.0934
\end{array}
\right)$$
It has been numerically checked that for any $n$ (at least for $n \leq 1000$)
$$f(x_0) \times f''(x_0) > 0$$ which implies, by Darboux theorem, that the solution will always be reached without any overshoot. However, since this product decreases quite fast when $n$ increases, we can expect more iterations for large $n$ (this is quite clear from the table).
I was hoping that, starting from $x_0$, Halley or Housholder methods would be leading to better $x_1$; this is not the case, the improvement being systematically quite marginal.
However, there is a way to avoid iterations (this means "almost" exact solution). Build around $x_0$ the $[1,m]$ Padé approximant which will write
$$f(x) \sim \frac {f(x_0)+a^{(m)}(x-x_0) } {1+\sum_{i=1}^m b_i^{(m)} (x-x_0)^i}$$ which is at least $O\big((x-x_0)^{m+1} \big)$; all required coefficients are defined by the function and derivatives values at $x=x_0$. This gives
$$x_{(m)}=x_0-\frac{f(x_0) } {a^{(m)} }$$ This has been done for this problem with $m=1,2,3$ and the results are better and better.
Best Answer
By http://dlmf.nist.gov/10.14.E4 and the known bound $(n/e)^n < n!$, we have $$ J_n (x) \le \frac{1}{{n!}}\left( {\frac{x}{2}} \right)^n \le \left( {\frac{e}{n}} \right)^n \left( {\frac{x}{2}} \right)^n = \left( {\frac{{ex}}{{2n}}} \right)^n $$ for $n\in \mathbb{N}$ and $x>0$. For the first inequality, see page 49 in G. N. Watson's book A Treatise on the Theory of Bessel Functions.