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I am having some trouble in rewriting a sum whose upper limit is given in terms of a floor function $\lfloor \cdot \rfloor$. The task is to prove that both sides of the following expression coincide:

$$ C_1(\nu)\sum_{n=\nu+1}\sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} \frac{\beta^{2n}}{g(n,k)}\binom{2n}{2\nu+1}+C_2\sum_{n=\nu+1}\sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} \frac{\beta^{2n}}{g(n,k)}\binom{2n}{2\nu+2}\overset{!}{=}\\ C_3\sum_{n=\nu+1}\sum_{k=0}^{\lfloor \frac{n+1}{2} \rfloor} \frac{\beta^{2n}}{g(n+1,k)}\binom{2n+2}{2\nu+3}, $$

I am over-simplifying my expression, for instance I am hiding many terms in the constants $C_2, C_3$, and in the functions $C_1(\nu), g(n,k)$. (This terms have no importance in my question).

My question is the following, in order to see whether both sides of my equation coincide I want to write the k-sum in the RHS in terms of $\lfloor \frac{n}{2}\rfloor$ instead of $\lfloor\frac{n+1}{2}\rfloor$. Do you have any suggestions to be able to rewrite the sum? I have tried to break the brackets by considering the odd and even values but I haven't been successful.

Thanks in advance.

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  • $\begingroup$ Is not $$C_3\sum_{n=\nu+1}\sum_{k=0}^{\lfloor \frac{n+1}{2} \rfloor} \frac{\beta^{2n}}{g(n+1,k)}\binom{2n+2}{2\nu+3}=C_3\sum_{n=\nu+1}\sum_{k=-1}^{\lfloor \frac{n}{2} \rfloor} \frac{\beta^{2n}}{g(n+1,k+1)}\binom{2n+2}{2\nu+3}?$$Also, you don't really need the floor functions in the upper limits of the sums since the sums will terminate at the same points with or without them. $\endgroup$
    – Steven Clark
    Commented May 4, 2023 at 20:49
  • $\begingroup$ The notation $\sum_{n = \nu + 1}( \dots)$ does not make sense. $\endgroup$
    – JBL
    Commented May 5, 2023 at 21:58

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