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I am evaluating using Mathematica, the double sum $\sum_{u=0}^\infty \lbrace \sum_{k= u+1}^{u+y}[\dfrac{(1-a)}{4} (3/4)^k + 3a[(\dfrac{1}{2})^{k-1} - (\dfrac{3}{4})^{k-1} ]\rbrace $, where $'a' $ is a constant and $a>0$ and $y$ is a parameter taking values in $[1,\infty)$.

I am having trouble evaluating this sum for very large values of $y$ due to recursion exceeding a specified value. I need to also find the value of this sum as $y \rightarrow \infty$. It is taking endless time even for $y= 5000$ and I had to abort the program.

I request suggestions on how I may resolve this problem in Mathematica.

Thanks in advance.

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    $\begingroup$ This FullSimplify@ Sum[Sum[(1 - a)/4 (3/4)^k + 3 a ((1/2)^(k - 1) - (3/4)^(k - 1)), {k, u + 1, u + y}], {u, 0, Infinity}] gives an answer analytically in V.13.1 $\endgroup$
    – bmf
    Aug 25, 2022 at 15:39

1 Answer 1

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Clear["Global`*"]

sum[a_, y_] = 
 FullSimplify@
  Sum[(1 - a)/4 (3/4)^k + 3 a ((1/2)^(k - 1) - (3/4)^(k - 1)), {u, 0, 
    Infinity}, {k, u + 1, u + y}]

(* -3 4^-y (3^y - 4^y + (2^(2 + y) - 17 3^y + 13 4^y) a) *)

The limit as y approaches infinity

Limit[sum[a, y], y -> Infinity]

(* 3 - 39 a *)

Comparing with y == 5000

N[sum[a, 5000], 100] // Expand

enter image description here

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