Eulers formula with an infinite series? Alright so this is a real life problem and not just a homework thing. Ive borrowed money from a family member $16323 \rm dkk$ to be exact. Im borrowing this money for $211$ days and im borrowing it from and account with an interest rate of $0.85$%. My calculations says its : $$16323+16323\cdot0.0085e^{211/365} = 16570.33 \rm dkk$$. But then my mother who is educated in this sort of stuff says that you calculate interest rate of the interest rate, and interest rate on top of that and then an infinite cascade similar to a geometric series. Im saying use Eulers formula to do this since i dont wanna rip anybody off but shes not doing that. How can i caluclate this correctly with the finite series of interest rates?

Let     


*

*$P$=Principal (the amount of loan you took)=$16323$    

*$r$=Interest Rate p.a.$=0.85\%=0.0085$  

*$n$=Number of days the loan is outstanding = $211$

*$d$=Number of days in a year = $365$


Three methods:
(1) Daily rest   (with daily compounding)
Total amount to be repaid*:
$$P\left(1+\frac rd\right)^n=16323\left(1+\frac {0.00085}{365}\right)^{211}=16403.4$$
(2) Yearly rest   (interpolated for intra-year period)
Total amount to be repaid*:
$$P\left(1+\frac {rn}d\right)=16323\left(1+\frac {(0.00085)(211)}{365}\right)=16403.2$$
(3) Continuous Compounding   (Exponential)
Total amount to be repaid*:
$$P\exp\left(r\left(\frac {n}{d}\right)\right)=16323\;\exp\left((0.00085)\frac{211}{365}\right)=16403.4$$
(4) Intra-period Compounding
Total amount to be repaid*:
$$P\left(1+r\right)^{n/d}=16323(1.00085)^{211/365}=16403.1$$
*assuming no repayment of either principal or interest in the interim period
From above the amount to be repaid is approximately the same, i.e. $\approx 16403$. 

NB
If $n=d$, results from methods $(2)$ and $(4)$ are the same. 
The notation $\exp(x)$ means the same as $e^x$. 
Note that if you chop up a year into $m$ periods, the interest repayment is 
$$P\left[\left(1+\frac rm\right) ^m\right]^\frac nd$$
Setting $m=d$ gives the formula in method $(1)$ above. 
Taking the limit as $m\to \infty$ gives the formula in method $(3)$ above.
$$\lim_{m\to \infty}P\left[\left(1+\frac rm\right) ^m\right]^{n/d}
=P\left[\underbrace{\lim_{m\to \infty}\left(1+\frac rm\right) ^m}_{=e}\right]^{n/d}=Pe^{rn/d}$$
