I need to explain the convergence of $\sum\limits_{k=0}^{\infty}\frac{x^k}{k!}$
So my working out I used the ratio test and I got up to $\sum\limits_{k=0}^{\infty}\frac{a}{k}+1$
Since $\sum\limits_{k=0}^{\infty}\frac{1}{k}+1=0$
Then $\sum\limits_{k=0}^{\infty}\frac{a}{k}+1=0$ and is convergent.
Is this correct?
The ratio of successive terms is
$$\frac{\dfrac{x^{k+1}}{(k+1)!}}{\dfrac{x^k}{k!}}=\frac x{k+1}.$$
For all $k>x$,
$$\frac x{k+1}<\frac k{k+1}=r<1.$$
So the tail of the sumation is dominated by a geometric series of ratio $r<1$, which converges. (If you prefer, $\forall i>0,T_{k+i}<r^iT_k$.)
By ratio test
$\forall x\in \mathbb R $
$$\lim_{k\to+\infty}\frac{ \frac{|x|^{k+1}}{(k+1)!} }{ \frac{|x|^k}{k! }}=$$
$$\lim_{k\to+\infty}\frac{|x|}{k+1}=0<1$$
$\implies \forall x\in \mathbb R\;\; \sum \frac{x^k}{k!}$ convergent.
There have already been posted solutions that apply the ratio test. I thought, therefore, that it might be instructive to present a solution that applies the root test.
Recalling that $k!\ge (k/2)^{k/2}$, and applying the squeeze theorem reveals
$$\sqrt[k]{\left|\frac{x^k}{k!}\right|}\le \frac{|x|}{\sqrt{k/2}}\to 0\,\,\text{as}\,\,k\to \infty$$
Hence, we see from the root test that the series converges with a radius of convergence $\infty$.
