Simplifying fraction of infinite series There is $\frac{1+\left( jx\right) ^{-1/5}\sum\limits_{k=2}^{\infty }\frac{%
\left( jx\right) ^{k/5}}{k!}}{1+\left( jx\right) ^{1/5}%
+\sum\limits_{k=2}^{\infty }\frac{\left( jx\right) ^{k/5}}{k!}}$ 
(where $x\epsilon \mathbb{R}$, $x>10$ and $j=\sqrt{-1}$).
It is $\lim\limits_{k\rightarrow \infty }\frac{\frac{\left( jx\right) %
^{\left( k\right) /5}}{k!}}{\frac{\left( jx\right) ^{\left( k-1\right) /5}}{%
\left( k-1\right) !}}=\lim\limits_{k\rightarrow \infty }\frac{1}{k}=0$, 
i.e. the sum converges. 
Is it ok (why or why not?) to approximate (in first order) 
enumerator and denominator and say 
$\frac{1+\left( jx\right) ^{-1/5}\sum\limits_{k=2}^{\infty }\frac{\left( jx\right) ^{k/5}}{k!}}{%
1+\left( jx\right) ^{1/5}+\sum\limits_{k=2}^{\infty }\frac{\left( jx\right) %
^{k/5}}{k!}}\approx \frac{1}{1+\left( jx\right) ^{1/5}}$ ?
And how can I simplify 
$\frac{1+\left( jx\right) ^{-1/5}\sum\limits_{k=2}^{\infty }\frac{%
\left( jx\right) ^{k/5}}{k!}}{1+\left( jx\right) ^{1/5}+%
\sum\limits_{k=2}^{\infty }\frac{\left( jx\right) ^{k/5}}{k!}}$ ?

We can use the power series representation of the exponential function
\begin{align*}
\exp(x)=\sum_{k=0}^\infty \frac{x^k}{k!}\qquad\qquad\qquad x\in\mathbb{C}
\end{align*}
in order to simplify the expression. We get
\begin{align*}
\sum_{k=2}^\infty \frac{(jx)^{(k/5)}}{k!}&=\sum_{k=2}^\infty \frac{\left((jx)^\frac{1}{5}\right)^k}{k!}
=\exp\left((jx)^\frac{1}{5}\right)-1-(jx)^{\frac{1}{5}}\tag{1}
\end{align*}

We obtain with the help of (1):
  \begin{align*}
\frac{1+(jx)^{-\frac{1}{5}}\sum_{k=2}^\infty \frac{(jx)^{(k/5)}}{k!}}
{1+(jx)^{\frac{1}{5}}+\sum_{k=2}^\infty \frac{(jx)^{(k/5)}}{k!}}
&=\frac{1+(jx)^{-\frac{1}{5}}\left(\exp\left((jx)^\frac{1}{5}\right)-1-(jx)^{\frac{1}{5}}\right)}{1+(jx)^{\frac{1}{5}}+\left(\exp\left((jx)^\frac{1}{5}\right)-1-(jx)^{\frac{1}{5}}\right)}\\
&=\frac{(jx)^{-\frac{1}{5}}\left(\exp\left((jx)^\frac{1}{5}\right)-1\right)}{\exp\left((jx)^\frac{1}{5}\right)}\\
&=(jx)^{-\frac{1}{5}}\left(1-\exp\left(-(jx)^{\frac{1}{5}}\right)\right)\tag{2}\\
\end{align*}

