Determine (or evaluate) the sum of the series $\sum_{j=0}^\infty (-1)^j \frac{\frac{3}{2}}{\frac{3}{2}+j} \frac{x^{2j}}{(2j)!}, \ \ \ x\in\mathbb R$. Determine (or evaluate) the sum of the series
$$\sum_{j=0}^\infty (-1)^j \frac{\frac{3}{2}}{\frac{3}{2}+j} \frac{x^{2j}}{(2j)!}, \ \ \ x\in\mathbb R$$
and
$$\sum_{j=0}^\infty (-1)^j \frac{\frac{5}{2}}{\frac{5}{2}+j} \frac{x^{2j}}{(2j)!}, \ \ \ x\in\mathbb R$$
The best that I have managed to do is to notice that
$$\left|\frac{\frac{1}{2}+k}{\frac{1}{2}+k+j}\right|\leq 1, \ \ \  \text{for all}\ j,k\in\mathbb N.$$
Then the sum is estimated from above by an hyperbolic cosine.
Thanks in advance.

For the first problem (assuming $x\neq0$),
$$\sum\limits_{j=0}^{\infty}(-1)^j\frac{\frac{3}{2}}{\frac{3}{2}+j}\frac{x^{2j}}{(2j)!}$$
Define $f(x)=\sum\limits_{j=0}^{\infty}(-1)^j\frac{1}{2j+3}\frac{x^{2j+3}}{(2j)!}$. Then, the given sum is equal to $3f(x)/x^3$ for all $x\neq0$.
$$f'(x)=\sum\limits_{j=0}^{\infty}(-1)^j\frac{x^{2j+2}}{(2j)!}=x^2\cos x$$
Integrating by parts, we have $f(x)=x^2\sin x + 2x\cos x - 2\sin x + c$ for some constant $c$.
So, we have 
$$\sum\limits_{j=0}^{\infty}(-1)^j\frac{\frac{3}{2}}{\frac{3}{2}+j}\frac{x^{2j}}{(2j)!}=\frac{3(x^2\sin x + 2x\cos x - 2\sin x + c)}{x^3}$$
As $x$ goes to $0$, LHS goes to $1$. In order for the limit to exist on the RHS, we must have $c=0$. We can then evaluate the limit of the RHS by L'Hospital's rule to verify that it equals $1$. Hence, for $x\neq 0$, sum of the series is
$$\frac{3(x^2\sin x + 2x\cos x - 2\sin x)}{x^3}$$
Proceeding in exactly the same manner, the sum for the second problem evaluates to (for $x\neq 0$):
$$\frac{5(x^4\sin x +4x^3\cos x-12x^2\sin x-24x\cos x+24\sin x)}{x^5}$$
