# Substitution rules not working

I have this summation, $$S$$,...

\begin{alignat}{2}% & P(0) = (1 - A) \notag \\ & P(1) = (1 - A)(e^A - 1) \\ & P(i + 1) = \frac{1}{P(0,A)}\Bigl\{P(i) - [P(0) + P(1)] P(i,A) \\ & \quad \quad-\sum_{\nu=2}^{i} P(\nu) \cdot P(i - \nu + 1, A) \Bigr\}. \label{md1ssprob} \end{alignat}

$$S = \sum _{k=0}^{\infty } \left(1-\sum _{i=r+1}^{2 r+1} \binom{2 r+1}{i} \left(1-\frac{P(k,\lambda )}{\sum _{j=0}^k P(j,\lambda )}\right)^i \left(1-\left(1-\frac{P(k,\lambda )}{\sum _{j=0}^k P(j,\lambda )}\right)\right)^{2r+1-i }\right)$$

and I want to be able to substitute for some arbitrary value $$r$$, but it doesn't seem to want to work. It only works when I manually enter the values for the summation.

Here is the code I have (that works if I manually put in the $$r$$ value).

Pn[0, \[Lambda]_] := 1 - \[Lambda];
Pn = (1 - \[Lambda])*(E^\[Lambda] - 1);
Pn[n_, \[Lambda]_] := (1 - \[Lambda])*
Sum[E^(j*\[Lambda])*(-1)^(n -
j)*(((j*\[Lambda] + n - j)*(j*\[Lambda])^(n - j - 1))/(n -
j)!), {j, 0, n}];
Sum[1 - Sum[
Binomial[2*r + 1,
i]*(1 - Pn[k, \[Lambda]]/Sum[Pn[j, \[Lambda]], {j, 0, k}])^
i*(1 - (1 -
Pn[k, \[Lambda]]/Sum[Pn[j, \[Lambda]], {j, 0, k}]))^(2*r +
1 - i),
{i, r + 1, 2*r + 1}], {k, 0, 35}] /. {\[Lambda] ->
0.5, r -> 101}



Why does it do this? Why can't I just substitute some $$r$$ value using the $$\longrightarrow$$ rule?

Also, in the code I only sum to 35. The reason is that if I put the symbol $$\infty$$ in, it takes forever and never completes. Also, it overflows sometimes if the number is too high. If you have any thoughts on that as well, it would be appreciated.

• The target for defining a function is to give is a set, pair of values as the input. For me the copy'n'paste code works with Pn[101, 0.5]. Mathematica runs in this way used faster because the values are taken earlier before the symbolic sum has been created. In my machine You interchanged {r,[Lambda]} in the rules and did not work properly with the braces. Mathematica creates a signature for the functions to identify them in more detail in this. Think of using rules in this cases a working in three implicit step, two symbolic calc and the numerical instead one. May 8 at 17:25

Maybe like this:

 ClearAll["*"]; Remove["*"];(* Clears All*)

Pn[0, \[Lambda]_] := 1 - \[Lambda];
Pn[1, \[Lambda]_] := (1 - \[Lambda])*(E^\[Lambda] - 1);
Pn[n_, \[Lambda]_] := (1 - \[Lambda])*Sum[E^(j*\[Lambda])*(-1)^(n - j)*(((j*\[Lambda] + n - j)*(j*\[Lambda])^(n - j - 1))/(n - j)!), {j, 0, n}];

f[\[Lambda]_, r_] := Sum[1 - Sum[Binomial[2*r + 1, i]*(1 - Pn[k, \[Lambda]]/Sum[Pn[j, \[Lambda]], {j, 0, k}])^i*(1 - (1 - Pn[k, \[Lambda]]/Sum[Pn[j, \[Lambda]], {j, 0, k}]))^(2*r + 1 - i), {i, r + 1, 2*r + 1}], {k, 0, 35}]

N[f[1/2, 101], 50]
(* 1.0010532880860907539154104616060531372235322012173 *)

f[1/2, 101] (*Large expression ! *)


For k->Infinity I doubt there's a closed form for the Sum.