Show that $\frac{(2k+h)^2}{\sqrt{h^2+k^2}}\sum_{n=0}^{\infty}\frac{(2k+h)^n}{(n+2)!} \rightarrow 0$ as $(h,k)\rightarrow 0.$

Question

To do this, I need to show two things:

$$ \begin{array}{lcl} \lim_{(h,k)\rightarrow (0,0)} \frac{(2k+h)^2}{\sqrt{h^2+k^2}}& = & 0 \quad &(1)&\\ \sum_{n=0}^{\infty}\frac{(2k+h)^n}{(n+2)!}& = & \text{Bounded} \quad &(2)&\\ \end{array} $$

(1) is not hard once I go over to polar coordinates. But how can I show (2)? Will the ratiotest suffice?

$$\left| \frac{(2k+h)^{n+1}}{((n+1)+2)!}\cdot\frac{(n+2)!}{(2k+h)^n} \right|=\left| \frac{2k+h}{n} \right|\rightarrow0, \ \text{as} \ n\rightarrow\infty.$$

But this only means that the series is absolutely convergent for constant $h,k$. In my limit $(h,k)$ tends to the origin.

Answer 1

Ratio test is ok, as an alternative note simply that that $\sum_{n=0}^{\infty}\frac{(2k+h)^n}{(n+2)!}$ converges absolutely by comparison test with $\sum_{n=0}^{\infty}\frac{1}{(n+2)!}$ indeed

$$\frac{\frac{ |2k+h|^n }{(n+2)!}}{\frac{1}{(n+2)!}}=|2k+h|^n\to0$$

Answer 2

Hint

$$\sum_{n=0}^{\infty}\frac{(2k+h)^n}{(n+2)!}=\sum_{n=0}^{\infty}\frac{x^n}{(n+2)!}=\frac 1{x^2}\sum_{n=0}^{\infty}\frac{x^{n+2}}{(n+2)!}$$

Now, think about the expansion od $e^x$.

I am sure that you can take it from here.