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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?
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  • $\begingroup$ Welcome to MSE. Please use MathJax to format your posts. The easier your questions are to read, the better the response you will get. $\endgroup$ – saulspatz Jun 19 '20 at 13:46
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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 $.

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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 $t<t_{n+1}<t_n$, and remember that a monotonically decreasing sequence (not series) that is bounded below converges.

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  • $\begingroup$ 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 $\endgroup$ – Andrea Ciufo Jun 19 '20 at 14:07
  • $\begingroup$ 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 $\endgroup$ – Andrea Ciufo Jun 19 '20 at 14:09
  • $\begingroup$ Take the limit as $n\to\infty$ on both sides. Then solve for $t$. $\endgroup$ – saulspatz Jun 19 '20 at 15:31

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