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If you have a sequence which is not geometric or arithmetic or arithmetico–geometric. Is there any methodology to follow in order to have a formula for its sum ?

Take for example the following sequence: $\{0.9^{\frac{1}{2}(n-i+1)(i+n)}\}_{i=1}^n$. It is not a geometric or an arithmetic progression. I don't see how to split it into sums of sequences which are arithmetic or geometric. Is there any hints I can get to proceed with writing a formula for this sum ?

$$S_n = \sum_{i=1}^n 0.9^{\frac{1}{2}(n-i+1)(i+n)}$$

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I hope you’ve played with this, and noticed:
1.$\quad$It’s not a sum, it’s many sums, and each sum is finite.
2.$\quad$The base, $0.9$ in this case, plays no particular role, so that you can use any base $r$.
3.$\quad$The first few values are \begin{align} S_0&=0\\ S_1&=r\\ S_2&=r^3+r^2\\ S_3&=r^6+r^5+r^3\\ S^4&=r^{10}+r^9+r^7+r^4\\ S_5&=r^{15}+r^{14}+r^{12}+r^9+r^5\\ S_n&=r^n(S_{n-1}+1) \end{align} I see no way of getting a closed-form expression for $S_n$, a polynomial in $r$ of degree $\frac12(n^2+n)$, and most certainly not a numerical value once you evaluate $r$ to, in your case, $r=0.9\,$.

I do wonder where or how you came across this—without context, it seems a most unnatural problem.

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  • $\begingroup$ Thank you. To answer your question, I have edited my question to add some context from where/how I came across this. Any comments ? $\endgroup$
    – ForumsDZ ForumsDZ
    Commented Sep 17, 2017 at 19:22
  • $\begingroup$ Also, is there any way to prove that $\forall n' > n > 5$, we have $S_{n'} \leq S_n$ ? $\endgroup$
    – ForumsDZ ForumsDZ
    Commented Sep 17, 2017 at 20:04
  • $\begingroup$ The answer to your question on inequalities might depend on the value of $r$—after all, if $r=1$, it’s never true. $\endgroup$
    – Lubin
    Commented Sep 17, 2017 at 23:37
  • $\begingroup$ Probably is possible — but I didn’t see it immediately. $\endgroup$
    – Lubin
    Commented Sep 18, 2017 at 13:13

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