Answer:
Following are the given series for all x:
Step-by-step explanation:
Given equation:
[tex]\bold{J_0(x)=\sum_{n=0}^{\infty}\frac{((-1)^{n}(x^{2n}))}{(2^{2n})(n!)^2}}\\[/tex]
Let the value a so, the value of [tex]a_n[/tex] and the value of [tex]a_(n+1)[/tex]is:
[tex]\to a_n=\frac{(-1)^2n x^{2n}}{2^{2n}(n!)^2}[/tex]
[tex]\to a_{(n+1)}=\frac{(-1)^{n+1} x^{2(n+1)}}{2^{2(n+1)}((n+1))!^2}[/tex]
To calculates its series we divide the above value:
[tex]\left | \frac{a_(n+1)}{a_n}\right |= \frac{\frac{(-1)^{n+1} x^{2(n+1)}}{2^{2(n+1)}((n+1))!^2}}{\frac{(-1)^2n x^{2n}}{2^{2n}(n!)^2}}\\\\[/tex]
[tex]= \left | \frac{(-1)^{n+1} x^{2(n+1)}}{2^{2(n+1)}((n+1))!^2} \cdot \frac {2^{2n}(n!)^2}{(-1)^2n x^{2n}} \right |[/tex]
[tex]= \left | \frac{ x^{2n+2}}{2^{2n+2}(n+1)!^2} \cdot \frac {2^{2n}(n!)^2}{x^{2n}} \right |[/tex]
[tex]= \left | \frac{ x^{2n+2}}{2^{2n+2}(n+1)^2 (n!)^2} \cdot \frac {2^{2n}(n!)^2}{x^{2n}} \right |\\\\= \left | \frac{x^{2n}\cdot x^2}{2^{2n} \cdot 2^2(n+1)^2 (n!)^2} \cdot \frac {2^{2n}(n!)^2}{x^{2n}} \right |\\\\[/tex]
[tex]= \frac{x^2}{2^2(n+1)^2}\longrightarrow 0 <1[/tex] for all x
The final value of the converges series for all x.
