# Proof of convergence for Customer Retention Rate Series

I tried to practice and see old notes of Calculus 1, but I can't still find out the reason why my series converge.

Retention Rate is a ratio that defines how many customers will shop again after the first purchase in a fixed period of time.

The starting conditions are: 100 customers at t0 Retention Rate is =0.4 New customers on t1=10 and this value is a constant for t2,...tn

So we have this series:

t0=100

t1=100*0.4+10

t2=(100*0.4+10)*0.4+10

...

tn=Something*0.4+10

1. Why Converge?
2. How to calculate the final value fixed Retention rate, New Customers at each time interval and starting base?
• Welcome to MSE. Please use MathJax to format your posts. The easier your questions are to read, the better the response you will get. – saulspatz Jun 19 '20 at 13:46

More generally, if $$t_{n+1} =at_n+b$$, then $$\dfrac{t_{n+1}}{a^{n+1}} =\dfrac{at_n+b}{a^{n+1}} =\dfrac{t_n}{a^{n}}+\dfrac{b}{a^{n+1}}$$.

Letting $$s_n =\dfrac{t_n}{a^{n}}$$, $$s_{n+1} =s_n+\dfrac{b}{a^{n+1}}$$ or $$s_{n+1}-s_n =\dfrac{b}{a^{n+1}}$$.

Summing from $$1$$ to $$m-1$$, $$\sum_{n=1}^{m-1}(s_{n+1}-s_n) =\sum_{n=1}^{m-1}\dfrac{b}{a^{n+1}}$$ or $$s_m-s_1 =\sum_{n=1}^{m-1}\dfrac{b}{a^{n+1}}$$ or $$\dfrac{t_m}{a^m} =\dfrac{t_1}{a} +\sum_{n=1}^{m-1}ba^{-n-1} =\dfrac{t_1}{a} +\sum_{n=2}^{m}ba^{-n}$$ or

$$\begin{array}\\ t_m &=a^{m-1}t_1 +a^m\sum_{n=2}^{m}ba^{-n}\\ &=a^{m-1}t_1 +b\sum_{n=2}^{m}a^{n-m}\\ &=a^{m-1}t_1 +b\sum_{n=0}^{m-2}a^{n}\\ &=a^{m-1}t_1 +b\dfrac{1-a^{m-1}}{1-a} \qquad\text{if } a \ne 1\\ \end{array}$$

If $$|a| < 1$$ then $$t_m \to \dfrac{b}{1-a}$$.

If $$|a| > 1$$ then $$\dfrac{t_m}{a^{m-1}} =t_1+b\dfrac{a^{-m+1}-1}{1-a} \to t_1+\dfrac{b}{a-1}$$.

If $$a = 1$$ then $$t_m = t_1+(m-1)b$$.

We have $$t_{n+1}=.4t_n+10$$ Suppose $$\lim_{n\to\infty}t_n=t$$. What must the value of $$t$$ be?

As for convergence, show that $$t, and remember that a monotonically decreasing sequence (not series) that is bounded below converges.

• I feel so dumb :) For the first question, the answer seems to be t because it's a costant. Based on that I would say that the answer is 50, but I think to be complitely wrong, empirically It's 16.6 – Andrea Ciufo Jun 19 '20 at 14:07
• In another example that I made to understand better. With t0=1000 Retention Rate=0.7 and Constant Value 1000, It converges to 3333. But I am very rude, I have done as a proof on Excel and I am trying to understand why – Andrea Ciufo Jun 19 '20 at 14:09
• Take the limit as $n\to\infty$ on both sides. Then solve for $t$. – saulspatz Jun 19 '20 at 15:31