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Have here a sum of which it can be split into several sums. How to do this from the sum notation itself at once ?

Note :I do not want to use the command "Apart" to first split the fraction and then put it together in the sum notation. That is too laborious.

S = Sum[1/(16*n^2 - 8*n - 3), {n, 1, Infinity}]

How to go to this form?

$\underset{n=1}{\overset{\infty }{\sum }}\left(\frac{1}{4 (4 n-3)}-\frac{1}{4 (4 n+1)}\right)$

EDIT( MMA can calculate this in one step this finding of the 1/4 value of the sum, or in steps ).

Its a telescoping sum $\left\{\frac{1}{4 (4 \text{n}-3)}\right\}_{\text{n=1..10}}=\left\{\frac{1}{4},\frac{1}{20},\frac{1}{36},\frac{1}{52},\frac{1}{68},\frac{1}{84},\frac{1}{100},\frac{1}{116},\frac{1}{132},\frac{1}{148},\ldots \right\} $

$\left\{-\frac{1}{4 (4 \text{n}+1)}\right\}_{\text{n=1..10}}=\left\{-\frac{1}{20},-\frac{1}{36},-\frac{1}{52},-\frac{1}{68},-\frac{1}{84},-\frac{1}{100},-\frac{1}{116},-\frac{1}{132},-\frac{1}{148},-\frac{1}{164},\ldots \right\}$

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    $\begingroup$ Once again, your question is unclear. What's the desired output? Is HoldForm@Sum[#, {n, 1, Infinity}] & /@ Apart[1/(16*n^2 - 8*n - 3)] desired? What do you mean by "first split the fraction and then put it together in the sum notation"? Is the output for display only, or will be used in subsequent programming? $\endgroup$
    – xzczd
    Mar 19, 2022 at 12:32
  • $\begingroup$ @xczd i must give this more attention : better explaning I added in the question the desired output , now you will get the idea $\endgroup$
    – janhardo
    Mar 19, 2022 at 12:51
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    $\begingroup$ Maybe: Inactivate[Sum[Apart[1/(16*n^2 - 8*n - 3)], {n, 1, Infinity}], Sum] ? $\endgroup$ Mar 19, 2022 at 12:52
  • $\begingroup$ Its correct your answer, and the two terms in the sum look similar : they are cancelling eachother out almost : a telescoping sum $\endgroup$
    – janhardo
    Mar 19, 2022 at 13:05
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    $\begingroup$ But it uses Apart. Your LateX expression is not correct. It has 4n terms in the numerator? $\endgroup$
    – Syed
    Mar 19, 2022 at 13:10

1 Answer 1

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Clear["Global`*"]

Sum[1/(16*n^2 - 8*n - 3), {n, 1, Infinity}]

(* 1/4 *)

Step-by-step

(S = Inactive[Sum][1/(16*n^2 - 8*n - 3), {n, 1, Infinity}]) // 
  TraditionalForm

enter image description here

(S = Inactive[Sum][
    1/(16*n^2 - 8*n - 3) // Apart, {n, 1, Infinity}]) // 
  TraditionalForm

enter image description here

(S = Inactive[Sum][#, {n, 1, Infinity}] & /@ S[[1]]) // 
  TraditionalForm

enter image description here

(S = (nmax = 10; 
   S /. Inactive[Sum][expr_, {var_, lb_, ub_}] :> 
     Inactive[Plus] @@ Append[Table[expr, {var, lb, nmax--}], …]))

enter image description here

S = (S /. … :> 0) // Activate

(* 1/4 *)
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  • $\begingroup$ Thanks, This is really step by step done. Was wondering how to split an expression embedded in a sum notation. $\endgroup$
    – janhardo
    Mar 19, 2022 at 16:25

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