 Solution

For a nonzero $x$, the absolute successive-term ratio in the exponential series is $|x|/(n+1)\to 0$, so the first series converges for every $x$. For the factorial series the ratio is $(n+1)|x|\to \infty$, so its terms fail to tend to zero for every $x\ne 0$.

The third series is sparse: its exponent is $n^2$, as in the PDF. If $t_n=(n!{)}^2|x{|}^{n^2}$, thenFor $|x|>1$, the terms grow without bound; at $|x|=1$ their moduli are $(n!{)}^2$, so they again fail the term test. Equivalently, the Cauchy-Hadamard theorem for the sparse coefficients uses $(n!{)}^{2/n^2}\to 1$, not $(n!{)}^{2/n}\to \infty$. Thus the three radii of convergence areZeros at non-square coefficient indices do not change the relevant limsup in the last root test. This illustrates the radius of convergence of a sparse power series.