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No solution from RSolve[ ] of two equations?

Posted 2 months ago
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enter image description here

I know

  1. S[n] is Sum[a[k], {k,1,n}]
  2. a[1]==1

I want to get the general solution of a[n] and S[n].

So I tried this one.enter image description here

But I can't get the answer. What can I do for this problem?

4 Replies

I do not understand what S_k...(*) means. Maybe this approach is helpful:

s[n_] := Sum[a[i], {i, 1, n}]
ss[n_] := Sum[(n - i + 1) a[i], {i, 1, n}]
(* your equation: *)
eqn[n_] := (n + 1) s[n + 1] == Log[2, n + 2] + ss[n]
nMax = 10;  (* as an example *)
vars = Table[a[n], {n, 1, nMax + 1}];
equations = Table[eqn[n], {n, 0, nMax}];
(* solutions: *)
Solve[equations, vars]

Though not rigorous according to this it seems that $$ a[n]= -\frac{\log(n) - \log(n+1)}{n \log(2)} $$ is the general solution.

Posted 2 months ago

Oh. Thank you very much. Your approach is helpful for me.

@Park Seong Hyun Welcome to Wolfram Community! Please next time make sure you know the rules: https://wolfr.am/READ-1ST

The rules explain how to format your code properly. Posting code Images doesn't help other members to copy your code. Please EDIT your post and make sure code blocks start on a new paragraph and look framed and colored like this.

int = Integrate[1/(x^3 - 1), x];
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enter image description here

Looking at your equations eqn[ n ] (in Henriks notation). expanding them (by hand ) and and forming the difference eqn[ j + 1 ] - eqn [ j ] you arrive at

( j + 2 ) a[ j + 2 ] == Log[ ( j + 3 ) / ( j + 2 ) ] / Log[ 2 ]

and therefore

a[ j_] := Log[(1 + j)/j]/(j Log[2])
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