# Prepaying interest with set installment amount where extra payment is applied to principal

I believe this is a problem of computing the limit of a recursive sequence.

Let's say I have a mortgage of \$332950 and a 6% annual interest rate. The mortgage contract requires that interest for each year is prepaid for each year on Dec 31 in the amount of \$20,000. (Interest for 2016 is paid on Dec 31 2015) Any amount paid above the required interest is applied to the principal. Now, 6% of \$332950 is \$19977, NOT \$20000, so by default the contract requires paying more than just interest, and so some of that money will be applied to the principal. The problem is that if \$23 of the $20000 installment is applied to principal, then the the interest due is thereby decreased. If the interest due is decreased, then even more is applied to principal. I ran a 100 iterations of this cycle using a python script, and after 10 iterations, the amount applied to interest reaches a limit of \$24.4680851064. How can I compute this limit without using brute force?

If I understand the notation correctly, this is how to define the sequence:

Suppose $a_0 = 0$, and $$a_n = 20000 - ( ( 332950 - a_{n-1}) \cdot 0.06)$$ if $n \ge 1$.

How do I compute the limit as $n$ approaches $\infty$?

iteration   a
0           0
1           23
2           24.38
3           24.4628
4           24.467768
5           24.46806608
6           24.46808396
7           24.46808504
8           24.4680851
9           24.46808511
10          24.46808511
...         ...
100,000,000 24.46808511

• If $lim_{n\to +\infty}a_n=L$ then $\lim_{n\to +\infty}a_{n-1}=L$. – hamam_Abdallah Dec 21 '16 at 21:17
• Replace bothv$a_n$ and $a_{n-1}$ by $L$ and solve. you will find tge limit L. – hamam_Abdallah Dec 22 '16 at 9:05
• The value $24,46...$ you got is not the limit. to reach the limit you need more iterations. – hamam_Abdallah Dec 22 '16 at 17:58
• @SalahFatima But it doesn't make sense that the limit would be lower than $23 (the original amount applied to principal), since I assume it should be monotonic (only grow). – reynoldsnlp Dec 22 '16 at 18:04 • You made a mistake. it is$+0.06L$not$-$– hamam_Abdallah Dec 22 '16 at 19:36 ## 1 Answer Thanks to @SalahFatima for the comments. If I replace$a_n$and$a_{n-1}$with$L\$, then I have the following solution:

$$a_n = 20000 - ( ( 332950 - a_{n-1}) \cdot 0.06)$$ $$L = 20000 - ( ( 332950 - L) \cdot 0.06)$$ $$L = 20000 - 19977 + 0.06L$$ $$L = 23 + 0.06L$$ $$0.94L = 23$$ $$L = \frac{23}{0.94}$$ $$L = 24.4680851$$