Prove or disprove: The Cauchy-product of serieses $\sum_{k=0}^{\infty}\frac{1}{k!}$ and $\sum_{k=0}^{\infty}\frac{2^{k}}{k!}$ converges to $e^{2}$

Question

Prove or disprove: The Cauchy-product of serieses $\sum_{k=0}^{\infty}\frac{1}{k!}$ and $\sum_{k=0}^{\infty}\frac{2^{k}}{k!}$ converges to $e^{2}$.

I'm not really sure if it's done like that but I have started by multiplying both serieses with each other, then use ratio test on the product:

$$\sum_{k=0}^{\infty}\left(\frac{1}{k!}\cdot \frac{2^{k}}{k!} \right )$$

$\Rightarrow$

$$\lim_{k\rightarrow\infty}\left|\left(\frac{\frac{2^{k+1}}{(k+1)!\cdot (k+1)!}}{\frac{2^{k}}{k!\cdot k!}} \right)\right |=\lim_{k\rightarrow\infty}\left|\left(\frac{2^{k+1}\cdot k!\cdot k!}{(k+1)!\cdot (k+1)! \cdot 2^{k}}\right)\right |$$

$$= \lim_{k\rightarrow\infty}\left|\left(\frac{2^{k}\cdot 2 \cdot k! \cdot k!}{k!\cdot (k+1)\cdot k!\cdot (k+1)\cdot 2^{k}}\right)\right |= \lim_{k\rightarrow\infty}\left(\frac{2}{(k+1)^{2}}\right)= 0$$

Thus the statement is false.

Answer 1

You are making a mistake from the beginning, the termwise product is not the Cauchy product. You should be looking at $$ \sum_{k=0}^\infty a_k \sum_{k=0}^\infty b_k = \sum_{k=0}^\infty c_k $$ with $$c_k = \sum_{\ell=0}^k a_\ell b_{k-\ell}.$$

In your case, this will lead to $$c_k = \sum_{\ell=0}^k \frac{1}{(k-\ell)!}\frac{2^{\ell}}{\ell!} = \frac{1}{k!}\sum_{\ell=0}^k \frac{k!}{(k-\ell)!\ell!}{2^{\ell}} = \frac{(1+2)^k}{k!} = \frac{3^k}{k!}$$ and the (Cauchy) product will then be $$ \sum_{k=0}^\infty \frac{3^k}{k!} = \cdots $$

Answer 2

Let $f_1(x) = \sum_k {1 \over k!} x^k = e^x$. Similarly, let $f_2(x) = \sum_k {1 \over k!} (2x)^k = e^{2x}$.

Let $f_3(x) = f_1(x) f_2(x) = e^{3x}$, then the power series coefficients of $f_3$ are given by the Cauchy product of the power series coefficients of $f_1,f_2$ and so the Cauchy product is given by $f_3(1) = e^3$.