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I'd like to bound the partial sum: $S(p,q,\alpha):=\sum\limits^{p+q-1}_{k=[\alpha(p+q)]}\{-\frac{qk}{p+q}\}-\sum\limits^{p+2q-1}_{[\alpha(p+2q)]}\{-\frac{qk}{p+2q}\}$.

Here $p,q$ are naturals, $0<\alpha<1$, while $[]$, $\{\}$ mean the integer/fractional parts. I guess there is no hope to compute this. But the numerics suggests: $S(p,q,\alpha)<\frac{1}{1-\alpha}$. I guess each of the sums can behave badly, but the bad things cancel out.

Though this bound is not the best possible, I'd be happy to see something of the type: $S(p,q,\alpha)<f(\alpha)+q\cdot g(\alpha)$. Where $f,g$ are some (reasonable) functions. (And the bound is asymptotically sharp.)

I'd like to have also some lower bound.

Actually this is just the simplest in the sequence of expressions. The next beast to bound is: $S(p_1,p_2,p_3,\alpha):=\sum\limits^{p_1+p_2+2p_3-1}_{k=[\alpha(p_1+p_2+2p_3)]}\{\frac{p_1k}{p_1+p_2+2p_3}\}-\sum\limits^{p_1+2p_2+3p_3-1}_{[\alpha(p_1+2p_2+3p_3)]}\{-\frac{p_2k}{p_1+2p_2+3p_3}\}+\sum\limits^{2p_1+3p_2+6p_3-1}_{[\alpha(2p_1+3p_2+6p_3)]}\Big(\{\frac{(p_1+p_2+3p_3)k}{2p_1+3p_2+6p_3}\}-\{\frac{(p_1+p_2+2p_3)k}{2p_1+3p_2+6p_3}\}\Big)$.

Again, some numerics suggests: $S(p_1,p_2,p_3,\alpha)<\frac{1}{1-\alpha}$. Any suggestions?

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  • $\begingroup$ Just to be clear, typically $S(p,q,\alpha)$ is negative, and can be arbitrarily negative, you just want to bound how positive it can become? $\endgroup$
    – Kevin O'Bryant
    Commented Mar 22, 2014 at 21:20
  • $\begingroup$ Isn't $S(28,18,1/2) = 336/23$, considerably larger than your numerics suggest? $\endgroup$
    – Kevin O'Bryant
    Commented Mar 22, 2014 at 21:52
  • $\begingroup$ Strange, I get S(28,18,1/2)=-9/2. Could you check your computation? Yes, I need at least the upper bound. (Though a lower bound would be also interesting.) $\endgroup$
    – Dmitry Kerner
    Commented Mar 25, 2014 at 11:47

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