Simplifying fraction of infinite series

Question

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 $\frac{1+\left( jx\right) ^{-1/5}\sum\limits_{k=2}^{\infty }\frac{% \left( jx\right) ^{k/5}}{k!}}{1+\left( jx\right) ^{1/5}+% \sum\limits_{k=2}^{\infty }\frac{\left( jx\right) ^{k/5}}{k!}}$ ?

Answer 1

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*}