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I have an equation $f(n+1)=f(n) + (1 − f(n))\cdot \frac{1}{2+}\cdot\frac{2}{3+}\cdot\frac{6}{7+}$ , $f(1)=\frac{2}{7}$. I can get a solution using WolframAlpha ($f(n)=\frac{n(n+7)}{2(n+1)(n+6)}$ ), but I wonder if it is solved analytically. In the materials I found, only simple cases with constant coefficients are described, or equations where everything conveniently simplifies. However, I can't seem to manage this here. I tried to find a generating function, but nothing worked out due to the arrangement of the coefficients. I would be grateful if someone could explain how to approach such equations or recommend suitable articles that describe this.

Upd:

Let $f(n)$ be denoted by $a_{n}$.

Then $a_{1} = \frac{2}{7}$.

Next, we solve the recurrence relation:

$ a_{n+1} = a_{n} + (1 - a_{n}) \cdot \frac{1}{2 + n} \cdot \frac{2}{3 + n} \cdot \frac{6}{7 + n} $

Let $ g_{n} = \frac{1}{2 + n} \cdot \frac{2}{3 + n} \cdot \frac{6}{7 + n} $.

Then:

$ a_{n+1} = a_{n} + (1 - a_{n}) \cdot g_{n} $

$ a_{n+1} = a_{n} \cdot (1 - g_{n}) + g_{n} $

Let $ f_{n} = 1 - g_{n} $.

We get:

$ a_{n+1} - f_{n}a_{n} = g_{n} $

$ \frac{a_{n+1}}{\prod_{k = 0}^{n} f_{k}} - \frac{a_{n}}{\prod_{k = 0}^{n-1} f_{k}} = \frac{g_{n}}{\prod_{k = 0}^{n} f_{k}} $

Let $ A_{n} = \frac{a_{n}}{\prod_{k = 0}^{n-1} f_{k}} $.

Then:

$ A_{n+1} - A_{n} = \frac{g_{n}}{\prod_{k = 0}^{n} f_{k}} $

Consider the sum:

$ \sum_{k=0}^{n-1} (A_{k+1} - A_{k}) = A_{n} - A_{0} = \sum_{k=0}^{n-1} \frac{g_{k}}{\prod_{m = 0}^{k} f_{m}} $

In our case, $ A_{0} = 0 $.

Then: $ \frac{a_{n}}{\prod_{k = 0}^{n-1} f_{k}} = \sum_{m=0}^{n-1} \frac{g_{m}}{\prod_{k = 0}^{m} f_{k}} $

Thus:

$ a_{n} = \left( \prod_{k = 0}^{n-1} f_{k} \right) \cdot \left( \sum_{m=0}^{n-1} \frac{g_{m}}{\prod_{k = 0}^{m} f_{k}} \right) $

Consider products of the form: $ \prod_{k = 0}^{n} f_{k} $

$ \prod_{k = 0}^{n} f_{k} = \prod_{k = 0}^{n} \left( 1 - \frac{1}{2 + k} \cdot \frac{2}{3 + k} \cdot \frac{6}{7 + k} \right) = \prod_{k = 0}^{n} \frac{(k+6)(k+5)(k+1)}{(k+2)(k+3)(k+7)} = \frac{(n+4)(n+5)}{2(n+2)(n+7)} $

Then:

$a_{n} = \frac{(n+3)(n+4)}{2(n+1)(n+6)} \cdot \sum_{m=0}^{n-1} \frac{\frac{12}{(2+m)(3+m)(7+m)}}{\frac{(m+4)(m+5)}{2(m+2)(m+7)}} = \frac{(n+3)(n+4)}{2(n+1)(n+6)} \cdot \sum_{m=0}^{n-1} \frac{24}{(m+3)(m+4)(m+5)} = \frac{(n+3)(n+4)}{2(n+1)(n+6)} \cdot \sum_{m=0}^{n-1} \left( \frac{12}{m+3} - \frac{24}{m+4} + \frac{12}{m+5} \right) = \frac{(n+3)(n+4)}{2(n+1)(n+6)} \cdot \frac{n(n+7)}{(n+3)(n+4)} = \frac{n(n+7)}{2(n+1)(n+6)} $

Of course, I got the answer, but what should be done in the general case if it is not possible to collapse sums and products using such a convenient method?

Trees
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  • Try Wikipedia https://en.m.wikipedia.org/wiki/Recurrence_relation. ‘Solving first-order non-homogeneous recurrence relations with variable coefficients’ – miracle173 Jun 30 '24 at 14:12
  • Thank you, the article indeed solves my type of equations, but I don't quite understand the transition to linear differential equations (including the notation nh). – Trees Jun 30 '24 at 15:11
  • You may follow this answer of mine : https://math.stackexchange.com/questions/4610074/recurrence-relation-in-reduction-formula-for-integral-sinnx/4610252#4610252, from the line $c_n = \alpha_nc_{n-1} + \beta_n$, with $\alpha_n = 1-\beta_n$ and $\beta_n = \frac{12}{(n+7)(n+3)(n+2)}$ in your case. – Abezhiko Jun 30 '24 at 15:26
  • Yes, I have already reached this form using the method from Wikipedia, but I need to arrive at an explicit formula (as specified in the question), which does not require the calculation of sums and products. – Trees Jun 30 '24 at 16:02
  • I did come to the answer after finding this form, but it seems it was only because the sums turned out to be telescoping. However, is there a method if the sum does not collapse so obviously? – Trees Jun 30 '24 at 16:42

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