The first 12 terms of the series are: 0.2x 0.08x^2, 0.048x^3, 0.0384x^4, 0.0384x^5, 0.04608x^6, 0.064512x^7, 0.1032192x^8, 0.18579456x^9, 0.37158912x^10, 0.817496064x^11, 1.961990554x^12, ..., 25510.82656x^20, ...
It's clear that the minimum value for the coefficient is 0.0384 which applies to the 4th and 5th terms. Beyond that the coefficient increases. The 20th term has a large coefficient, and the coefficient continues to increase thereafter. For convergence, the value of x^n must be such that (n!/5^n)x^n is progressively less than ((n-1)!/5^(n-1)x^(n-1). Therefore a[n]x^n<a[n-1]x^(n-1), a[n]x<a[n-1], and x<a[n-1]/a[n], where a[n] is the nth coefficient.
The ratio a[n]/a[n-1]=(n!/(n-1)!)/5=n/5, i.e., a[n]=(n/5)a[n-1]. Up to n=5 this is <1 but after that it is >1, so the radius of convergence>1 as n approaches infinity, implying divergence.
x<((n-1)!/n!)(5^n/5^(n-1)), x<5/n. If x>0, then But as n increases x is required to be smaller and smaller. As n approaches infinity, x will approach zero, so for the series to converge x=0 and the nth term of the series would be zero.
The question doesn't specifically mention the sum of the series, it just presents the series as a set of terms. If, however, the sum of the terms is implied, we can see already that for positive x, the sum of the terms diverges unconditionally, because the tendency for large n is that each term is bigger than its predecessor, except for the trivial case of x=0.
If x<0 it satisfies the requirement x<5/n because n>0. Consider 2 consecutive terms in the series:
a[n-1]x^(n-1), a[n]x^n. These will have opposite sign. If A[n] is the complete term A[n]=nxA[n-1]/5 and A[n]-A[n-1]=nxA[n-1]/5-A[n-1]=A[n-1](nx/5-1). As n becomes large this difference tends to A[n-1](nx/5) which is virtually the same as when terms are added, and divergence applies when x<0.