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Does the sum $\sum_{i=1}^{n}\frac{2^{\alpha_1+\ldots+\alpha_i}}{3^iv_1+\sum_{j=1}^{i}3^{j-1}2^{\alpha_1+\ldots+\alpha_n-\sum_{l>i-j}\alpha_l}}$ for a positive odd integer $v_1$ and a sequence of positive integers $\alpha_i$ has a limit?

Let us take for example the Collatz sequence $v_1=17$, $v_2=13$, $v_3=5$, $v_4=1$.

We can calculate $v_{n+1}=\frac{3^nv_1+\sum_{j=1}^{n}3^{j-1}2^{\alpha_1+\ldots+\alpha_n-\sum_{l>n-j}\alpha_l}}{2^{\alpha_1+\ldots+\alpha_n}}$.

For the above-given example $v_1=17$, $\alpha_1=2$, $\alpha_2=3$ and $\alpha_3=4$ finally $v_4$ results in $1$: $v_4=v_{3+1}=1$

Since we know that a product $\prod_{i=1}^{n}(1+\frac{1}{3v_i})$ converges towards a limit, if the sum $\sum_{i=1}^{n}(\frac{1}{3v_i})$ is limited. We found out empirically that the product for any Collatz sequence stays below 2.

The idea is to show that the sum $\sum_{i=1}^{n+1}(\frac{1}{3v_i})=\frac{1}{3v_1}+\sum_{i=1}^{n}\frac{1}{3v_{i+1}}$ converges.

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    If the first sum of your post is the last sum of your post (RHS part), shouldn't you have $3^{j-1}$ in the denominator's sum? also in $v_{n+1}$ formula you put $\alpha_i$ instead of $\alpha_n$. – Collag3n Mar 21 '20 at 14:01
  • You are completely right - I fixed this. Thank you very much for that hint! –  Mar 21 '20 at 14:34
  • Using instead of the Collatz case (k=3) the more simpler case k=1, such a proof should be straightforward and the sum becomes $\sum_{i=1}^{n}\frac{2^{\alpha_1+\ldots+\alpha_i}}{v_1+\sum_{j=1}^{i}2^{\alpha_1+\ldots+\alpha_n-\sum_{l>i-j}\alpha_l}}$. Assuming $\alpha_i=1$ maximizes the sum, that sum becomes much easier: https://math.stackexchange.com/questions/3573413/what-is-the-limit-of-the-sum-sum-i-1n-frac1b-i-for-the-decreasing-se –  Mar 21 '20 at 14:46

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