Can $\frac{e^{r(n+1)}-1}{e^r-1}$ be simplified?

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In coming up with an balance formula for investment accounts with repeated regular contributions using continuous compounding, regular contribution $ P $, and periodical interest rate $ r $ I arrived at:

$$ B_n = P + Pe^r + Pe^{2r} + ... + Pe^{nr} = \sum_{k=0}^{n} Pe^{kr}$$

Which, by the formula for closed form geometric series, decomposes into:

$$ B_n = P\frac{e^{r(n+1)}-1}{e^r-1} $$

(I'm assuming that $ r > 0 $, so $ e^r > 1 $, and I prefer $\frac{e^{r(n+1)}-1}{e^r-1}$ to $\frac{1-e^{r(n+1)}}{1-e^r}$)

The term $ \frac {e^{r(n+1)}-1}{e^r-1} $ has been driving me nuts because it doesn't seem to want to simplify, so it doesn't respond to the natural logarithm well if I want to solve for $ r $.

Is there any mathematical significance to $ \frac{e^{r(n+1)}-1}{e^r-1} $ ? Can it be simplified? Am I missing something?

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This is an XY question. The expression is already in its simplest form, and having a simple form is not a guarantee that you can invert it.

The problem of finding the interest rate for annuities is well-known to have no analytical solution, but can easily be solved numerically by Newton's iterations.