Is there a closed form expression for the following series? Is there a closed form expression for
$$
\sum_{1 \le k \le n}ke^{-t}\left(e^{k+\frac{1}{2k^3}}-e^{k-\frac{1}{2k^3}} \right)?
$$
 this series emerge in computation output of this system via convolution of input $u(t)$ and impulse response (inverse laplase transform of transfer function of system) 
$y(t)=e^{-t}*u(t)=\int e^{-(t-s)}u(s)ds$
part b of this problem
as bellow 
 A: This is just an extended comment and not necessarily an answer.
$$
\sum_{k} ke^{-t}\left(e^{k+\frac{1}{2k^3}}-e^{k-\frac{1}{2k^3}}\right)=e^{-t}\sum_k k\left(e^{k+\frac{1}{2k^3}}-e^{k-\frac{1}{2k^3}}\right).
$$
Since we have that $k+\dfrac{1}{2k^3}\ne k-\dfrac{1}{2k^3}$ for all values of $k$, it is impossible for the terms $\left(e^{k+\frac{1}{2k^3}}\right)_{1 \le k \le n}$ and $\left(e^{k-\frac{1}{2k^3}}\right)_{1\le k \le n}$ to add together as you cannot add two terms which have different exponents. Note that I am saying that we can't add them in the same way we can't add $x^a+x^b$. It does not mean that we literally cannot add them; it means that we cannot express them as a single thing (we must use the $+$ sign). In a more refined sense: There is no way of 'simplifying' this situation based on the way it is written above.
Thus, it's worth looking into the sum once it's separated:
$$
e^{-t}\sum_k k\left(e^{k+\frac{1}{2k^3}}-e^{k-\frac{1}{2k^3}}\right)=e^{-t}\sum_k ke^{k+\frac{1}{2k^3}}-\sum_k ke^{k-\frac{1}{2k^3}}.
$$
Sadly, we arrive at the exact same problem. The exponents are still different. The split only made this much more obvious.
There may be some benefit gained by playing around with the many different expressions of $e^x$ and $e$ itself (particularly the infinite series expansions), but such playing seems a bit like trying to ram a square peg into a hole that's not only round but also may not even exist. I hope someone wiser than myself can spread more light on this.

Some Cute Observations
Let us use the representation of $e^x$ as an infinite series:
$$
e^x=\sum_{n\ge 0}\frac{x^n}{n!}.
$$
Applying this to the expression $e^{k+\frac{1}{2k^3}}-e^{k-\frac{1}{2k^3}}$, we begin to make some cute observations:
\begin{align}
e^{k+\frac{1}{2k^3}}-e^{k-\frac{1}{2k^3}}&=\sum_{n\ge 0}\frac{\left(k+\frac{1}{2k^3}\right)^n}{n!}-\frac{\left(k-\frac{1}{2k^3}\right)^n}{n!}\\
&=\sum_{n \ge 0}\frac{\left(k+\frac{1}{2k^3}\right)^n-\left(k-\frac{1}{2k^3}\right)^n}{n!}\\
&=\sum_{n\ge 0}\frac{1}{n!}\left[\sum_{0 \le j \le n} {n\choose j}\left(k^{n-j}\left[\frac{1}{2k^3}\right]^j-k^{n-j}(-1)^j\left[\frac{1}{2k^3}\right]^j \right) \right]\,\, \text{via binomial theorem}\\
&=\sum_{n\ge 0}\frac{1}{n!}\left[\sum_{0 \le j \le n} {n\choose j}\frac{1}{2^j}\left(k^{n-4j}-(-1)^jk^{n-4j} \right) \right]\\
&=\sum_{n\ge 0}\frac{1}{n!}\left[\sum_{0 \le j \le n} {n\choose j}\frac{k^{n-4j}}{2^j}\left(1-(-1)^j \right) \right].
\end{align}
Thus, we have that the innermost summand is $0$ whenever $j$ is even. I don't know if this is even useful, but I found it interesting.
If we continue this and turn this into a triple summation expressed with a single sigma where the order of indices to be done are from last to first in descending order, we have:
$$
e^{-t}\sum_k k\left(e^{k+\frac{1}{2k^3}}-e^{k-\frac{1}{2k^3}}\right)=e^{-t}\sum_{\substack{k\\ n\ge 0\\ 0\le j\le n}}\frac{k^{n-4j+1}}{(n-j)!j!2^j}(1-(-1)^j).
$$
