Answer
The series converges for all $x$.
Thus, the interval of convergence is $\left( { - \infty ,\infty } \right)$.
Work Step by Step
Let ${a_n} = \dfrac{{{2^n}{x^n}}}{{n!}}$. We compute $\rho $ from the Ratio Test:
$\rho = \mathop {\lim }\limits_{n \to \infty } \left| {\dfrac{{{a_{n + 1}}}}{{{a_n}}}} \right| = \mathop {\lim }\limits_{n \to \infty } \left| {\dfrac{{{2^{n + 1}}}}{{\left( {n + 1} \right)!}}{x^{n + 1}}\cdot\dfrac{{n!}}{{{2^n}{x^n}}}} \right| = 2\left| x \right|\mathop {\lim }\limits_{n \to \infty } \left| {\dfrac{1}{{n + 1}}} \right| = 0$
Since $\rho = 0 \lt 1$, the series converges for all $x$.
Thus, the interval of convergence is $\left( { - \infty ,\infty } \right)$.
