Approximating the compond interest for a loan

I was about to give up with this problem when an idea came to my mind.

The approximate solution must not require solving even quadratic equations; so, if the approximation is a Padé approximant, it should be of degree $1$ in numerator and degree $m$ in denominator but, at the same time, the function to be approximated, around $i=0$, has to be $$\frac{i\,(i+1)^n}{(i+1)^n-1}-r$$ In such a way I should need to solve for $i$ equation $$\frac{1-n\, r}{n}+b_m\, i=0$$ which is simple except that coefficients $b_m$ become very quickly quite ugly.

For $m=2$, introducing as an intermediate variable $$y=\frac{2 (n r-1)}{n+1}$$ (which is $i_1$ in the question), I arrived to $$i\approx\frac{2y\left(6+(n-1)y\right)}{12+(n-1)(4-y)y}$$ which is, for sure, not as good as the solution given by a $[2,2]$ Padé approximant but better than the $i_3$.

Repeating the calculation for $n=180$ and $r=0.01$, this leads to $i=0.007319$ while the exact value is $i=0.007299$.

Now, I give up !

Edit (almost three years later)

In an naswer to this question, I reported results obtained meanwhile for the approximation $$i=y\,\frac{\sum_{k=0}^3 c_k\, y^k }{\sum_{k=0}^3 d_k\, y^k }\qquad \text{where}\qquad y=\frac{2 ( nr-1)}{n+1}$$ with $$\left( \begin{array}{cc} \color{red} {k} & \color{red} {c_k} \\ 0 & 15120 \left(n^2-20 n+1\right) \\ 1 & -1260 (n+2) \left(8 n^2+113 n-13\right) \\ 2 & -90 \left(90 n^4+355 n^3+942 n^2+705 n-148\right) \\ 3 & -4 (n+2) \left(194 n^4+388 n^3+687 n^2+493 n-142\right) \end{array} \right)$$ and $$\left( \begin{array}{cc} \color{red} {k} & \color{red} {d_k} \\ 0 & 15120 \left(n^2-20 n+1\right) \\ 1 & -3780 \left(2 n^3+57 n^2+57 n-8\right) \\ 2 & -60 \left(170 n^4+823 n^3+1560 n^2+508 n-145\right) \\ 3 & -(2 n+1) \left(874 n^4+1559 n^3+3336 n^2+1139 n-428\right) \end{array} \right)$$ Repeating the calculation for $n=180$ and $r=0.01$, this leads to $$i=\frac{338717216186170264}{46405051549088067895}\approx 0.00729915$$ while the exact value is $i\approx0.00729945$.

Here is my 1st order simple equation:

$$A = P (1/n + 0.6r)$$ when $r n < 1$

Assume your variable i = my variable r

Take a $1M loan example with 4% interest over 15 years, (r * n = 0.6 OR r * n < 1).

Interest loan equation = $7,495 per month (annually compounding)

1st order approx. = $7,556 per month

error = + 0.8%, not bad!

If you use month compounding, the error does get a little higher.

Here is some comparison results of A for a range of r and n values:

Here is error function:

Your boy seems very smart and you also seem like a maths wiz. I don't know why my approximation works (especially where the 0.6 constant come from?!?) - I am just an engineer and realized for low r*n values, the curves are reasonably linear and can be linearly interpolated by my simple equation