sum_(n=0)^oo n!(x/2)^n Find the values of x for which the series converges.

For the power series sum_(n=0)^oo n!(x/2)^n, we may apply Ratio Test.
In Ratio test, we determine the limit as:
lim_(n-gtoo)|a_(n+1)/a_n| = L
or
lim_(n-gtoo)|a_(n+1)*1/a_n| = L
 Then ,we follow the conditions:
a) L lt1 then the series converges absolutely
b) Lgt1 then the series diverges
c) L=1 or does not exist  then the test is inconclusive.The series may be divergent, conditionally convergent, or absolutely convergent.
The given power series sum_(n=0)^oo n!(x/2)^n has:
a_n =n!(x/2)^n
Then,
1/a_n=1/(n!)(2/x)^n
        =1/(n!)(2^n/x^n)
        =2^n/((n!)x^n)
a_(n+1) =(n+1)!(x/2)^(n+1)
            = (n+1)(n!) x^(n+1)/2^(n+1)
            = (n+1)(n!)(x^n*x)/(2^n*2)
            =((n+1)(n!)*x^n*x)/(2^n*2))
Applying the Ratio test on the power series, we set-up the limit as:
lim_(n-gtoo) |((n+1)(n!)*x^n*x)/(2^n*2)*2^n/((n!)x^n)|
Cancel out common factors: x^n, n! , and 2^n .
lim_(n-gtoo) |((n+1)x)/2|
Evaluate the limit.
lim_(n-gtoo) |((n+1)*x)/2| = |x/2|lim_(n-gtoo) |n+1|
                             = |x/2|* oo
                             = oo       
The limit value L= oo satisfies Lgt 1 for all x.
Therefore,  the power series sum_(n=0)^oo n!(x/2)^n  diverges for all x .
There is no interval for convergence.
Note: The radius of convergence is 0 . The x=0 satisfy the convergence at points where n!(x/2)^n=0 .