Motorized Wheel Chair Source

First, note that:

2 * 4 * 6 * ... * 2n
= 2(1 * 2 * 3 * ... * n)
= 2n!, by the definition of the factorial.

So we really have:

sum(n=1 to infinity) (n^2 * x^n)/(2n!).

From here, we can use the Ratio Test, which states that if |lim (n-->infinity) (a_n+1)/(a_n)| < 1, then sum(n=1 to infinity) a_n is convergent.

With a_n = (n^2 * x^n)/(2n!), then:

a_n+1 = [(n + 1)^2 * x^(n + 1)]/[2(n + 1)!].

So now:

(a_n+1)/(a_n) = {[(n + 1)^2 * x^(n + 1)]/[2(n + 1)!]}/[(n^2 * x^n)/(2n!)]
= [(n + 1)^2 * x^(n + 1) * 2n!]/[n^2 * x^n * 2(n + 1)!]
= [(1 + 1/n)^2 * x * 2/(n + 1)].

Taking the limit as n --> infinity gives (1 + 0) * x * 0 = 0. Therefore, the series converges for all x.

I hope this helps!