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According to the alternating series test a general term $b_n$ in the sum: $\sum\limits_{n=1}^{\infty}(-1)^{n+1}b_n$ should converge to zero: $\lim\limits_{n \rightarrow \infty} b_n = 0$

From reuns answer here we get such an alternating series with a quantity that should converge to zero:

$$\lim_{n\to\infty}b_n=\sum_{n=2}^{\infty}\sum_{\substack{a_1,a_2,\dots,a_k\geqslant 2\\a_1 a_2\cdots a_k= n<2^{k+1}}}\frac{\log(a_1)}{n^{1/2}\log^{3+\epsilon}(n)}=0$$

Is the sum above equal to zero?

I have the feeling there is something boldly wrong with this question because I have been editing the words several times before posting. What I am hoping for is that such a question/sum is undecidable in a Gödel sense because it involves a product of an infinite number of variables $$a_1 a_2 a_3 \cdots a_{k}$$ at the same time as the variable $n$ goes to $\infty$.

Edit 14.6.2021:

(*Mathematica start*)
"Part 3 start:"
Clear[a1, a2, a3, a3, a4];
Monitor[Sum[If[1 == n, ""[n], 0], {n, 2^0, 2^1 - 1}], n]
Monitor[Sum[
  Sum[If[a1 == n, ""[n], 0], {a1, 2, n}], {n, 2^1, 2^2 - 1}], n]
Monitor[Sum[
  Sum[Sum[If[a1*a2 == n, ""[n], 0], {a1, 2, n}], {a2, 2, n}], {n, 2^2,
    2^3 - 1}], n]
Monitor[Sum[
  Sum[Sum[Sum[If[a1*a2*a3 == n, ""[n], 0], {a1, 2, n}], {a2, 2, 
     n}], {a3, 2, n}], {n, 2^3, 2^4 - 1}], n]
Monitor[Sum[
  Sum[Sum[Sum[
     Sum[If[a1*a2*a3*a4 == n, ""[n], 0], {a1, 2, n}], {a2, 2, 
      n}], {a3, 2, n}], {a4, 2, n}], {n, 2^4, 2^5 - 1}], n]
"Part 3 end."
(*Mathematica end*)

gives:

""[1]

""[2] + ""[3]

""[4] + 2 ""[6]

""[8] + 3 ""[12]

""[16] + 4 ""[24]

which appears to be:

TableForm[Table[""[2^n] + n ""[3*2^(n - 1)], {n, 0, 4}]]

Edit 14.6.2021 later:

(*Mathematica start*)"Part 4 start:"
Clear[a1, a2, a3, a3, a4];
Monitor[Sum[If[1 == n, 1/(""[n]), 0], {n, 2^0, 2^1 - 1}], n]
Monitor[Sum[
  Sum[If[a1 == n, ""[a1]/(""[n]), 0], {a1, 2, n}], {n, 2^1, 
   2^2 - 1}], n]
Monitor[Sum[
  Sum[Sum[If[a1*a2 == n, ""[a1]/(""[n]), 0], {a1, 2, n}], {a2, 2, 
    n}], {n, 2^2, 2^3 - 1}], n]
Monitor[Sum[
  Sum[Sum[Sum[If[a1*a2*a3 == n, ""[a1]/(""[n]), 0], {a1, 2, n}], {a2, 
     2, n}], {a3, 2, n}], {n, 2^3, 2^4 - 1}], n]
Monitor[Sum[
  Sum[Sum[Sum[
     Sum[If[a1*a2*a3*a4 == n, ""[a1]/(""[n]), 0], {a1, 2, n}], {a2, 2,
       n}], {a3, 2, n}], {a4, 2, n}], {n, 2^4, 2^5 - 1}], n]
"Part 4 end."
(*Mathematica end*)

"Part 4 start:"

1/""[1]

2

""[2]/""[4] + ""[2]/""[6] + ""[3]/""[6]

""[2]/""[8] + (2 ""[2])/""[12] + ""[3]/""[12]

""[2]/""[16] + (3 ""[2])/""[24] + ""[3]/""[24]

"Part 4 end."

Mats Granvik
  • 7,614

1 Answers1

1

For any $n$ that contributes one of the $a_i$ is $3$ and the others are all $2$. So $n=3×2^{k-1}$.
There are $k$ sequences$\{a_1,...,a_k\}$ for that $n$, namely $(3,2,2,...,2), (2,3,2,...,2),..., (2,2,...,2,3)$. The sum of the $\log a_1$ is $\log n$.
I can't tell the role of $n$ in the equation because it is a dummy variable on the right-hand side but not on the left.

Empy2
  • 52,372