I am attempting to show that a sequence which I defined as $\{a_n\}$ where $a_n =\sum_{p=0}^{n}\frac{x^p}{p!}$ converges, in order to do so I have attempted the ratio test at which point I arrive at the term $a_n =\sum_{p=0}^{n}\frac{x^p}{p!}$ and $a_{n+1} =\sum_{p=0}^{n+1}\frac{x^p}{p!}$ then taking the ratio I arrive at $$\frac{a_{n+1}}{a_n}=\frac{\sum_{p=0}^{n}\frac{x^p}{p!}}{\sum_{p=0}^{n+1}\frac{x^p}{p!}}$$
Taking the limit of the above is proving to be difficult. I am attempting the ratio test is . If this prove impossible I would like to show that it converges in some other form.
I have proved that in the case that $x=1$ we have convergence because it is bounded above and is increasing, however I want to show it for this case or for all x? How can we show this? I have also thought about showing it is cauchy in some form which would imply that it is convergent.
The following are equivalent (by definition of series convergence):
(I) The sequence $\displaystyle\big(\sum_{p=0}^n \frac{x^p}{p!}\big)_{n=0}^{\infty}$ of partial sums $\displaystyle\sum_{p=0}^n \frac{x^p}{p!}$ of the infinite series $\displaystyle\sum_{p=0}^{\infty}\frac{x^p}{p!}$ converges.
(II) The infinite series $\displaystyle\sum_{p=0}^{\infty}\frac{x^p}{p!}$ converges.
The latter is implied by the following:
(III) The limit $\displaystyle\lim_{n\to\infty} \left|\frac{x^{n+1}/(n+1)!}{x^n/n!}\right| $ exists and is less than $1$.
The implication (III) $\Rightarrow$ (II) is an application of the ratio test:
Correct Ratio Test. An infinite series $\displaystyle\sum_{p=0}^{\infty} t_p$ converges if $\displaystyle\lim_{n\to\infty}|t_{n+1}/t_n|<1$.
You seem to be misinterpreting the ratio test as follows:
Wrong Ratio Test. An infinite sequence $(s_n)_{n=0}^{\infty}$ converges if $\displaystyle\lim_{n\to\infty}|s_{n+1}/s_n|<1$.
You seem to be applying this wrong ratio test with $\displaystyle s_n=\sum_{p=0}^n\frac{x^p}{p!}$.