Relationship between the sum of the marginal percent change and cumulative percent change? Let’s say I have a sequence denoted $a_n=a_0+kn$. The total percent increase from $n=0$ to $n$ is just $100\cdot\frac{kn}{a_0}$. 
However, let’s denote the sum of the marginal percent increase as $100\cdot \sum_{i=0}^{n-1} \frac{k}{a_0+ki}$. 
I’m wondering if there’s a connection between the total/overall percent increase and the sum of the marginal percent increases? I understand how percent increase is inherently multiplicative in nature, but I’m wondering if there’s a simple way to go from the result of the sum above to the cumulative percent increase. I tried computing a closed form of the sum first but I’m getting results with the polygamma function and I can’t figure out a conversion. 
Edit: to clarify, I’m wondering if it is possible without knowing $a_0$ or $k$. So you would be given $n$ and you would be given the sum of the marginal percentage increases. Can you then find purely from this information the cumulative percent increase?
 A: If $a_0 \gg kn$ the marginal percent increase does not change much, so is just the total percent increase divided by $n$, which is $100\frac k{a_0}$.  Otherwise the sum of the marginal percent increases will be smaller because the absolute increase stays the same and the base is increasing.  The next level of approximation would be to take the average of the marginal percent increases as $x$ and say the total percent increase is $(1+\frac x{100})^n$.  This should be accurate for $kn$ much closer to $a_0$.  Make up a spreadsheet and give it a try.
A: Given the values of $t$ (a real number) and $n$ (a positive integer),
and given the fact that
$$ t = 100\sum_{i=0}^{n-1}\frac{k}{a_0+ki} $$
where $a_0$ and $k$ are unknown,
you wish to find the value of $100\frac{kn}{a_0}.$
If you let $r=\frac{k}{a_0},$ then 
$$
\sum_{i=0}^{n-1}\frac{k}{a_0+ki}
=\sum_{i=0}^{n-1}\frac{r}{1+ir}
=r \sum_{i=0}^{n-1}\frac{1}{1+ir}.
$$
Define a function $f(r) = r\sum_{i=0}^{n-1}\frac{1}{1+ir}$; then

$f(0) = 0$, $f(r)$ is strictly increasing for $r \geq 0,$
and $\lim_{r\to+\infty} f(r) = +\infty,$
so if we restrict $a_0$ and $k$ so that $\frac{k}{a_0} \geq 0,$
then whenever $t \geq 0$
there is exactly one solution for the unknown $r$ in $100f(r) = t$
provided that $t \geq 0.$
You can then find the value of $100\frac{kn}{a_0}$
by using that value of $r$ in the formula
$$100\frac{kn}{a_0} = 100rn.$$
For all but a few possible values of $n,$ I do not believe there is a conventional closed-form solution of $100f(r) = t,$
so in general I would expect to have to solve for $r$ using numerical methods.
Also note that $f(r)$ is undefined at 
$r=-1,-\frac12,-\frac13,\ldots,-\frac{1}{n-1},$
that $f(r)$ tends to $\pm\infty$ near those values of $r,$
that $\lim_{r\to-\infty} f(r) = -\infty,$
but $f'(r) > 0$ whenever $f(r)$ is defined.
Graphing $y=f(r)$ according to those facts, it is clear that
if we allow solutions such that $r < 0,$
there are $n$ possible solutions of $100 f(r) = t$ for any real number $t.$
But if you assume that $a_0 > 0$ and $k \geq 0,$ you avoid this complication.
