I'm trying to solve the following summation (where C is some constant) but I'm stuck because of the inner most summation which is limited by $i\sqrt[2]{j}$ where i and j are the iterators of the outer two summations. $$ \sum\limits_{i=1}^N \sum\limits_{j=1}^i \sum\limits_{k=1}^{i\sqrt[2]{j}}C $$
If that darned third summation didn't exist, I would be able to solve this easily but I have no idea how to deal with that third summation. Any thoughts?
edit: $i, j, k\in\mathbb Z_{> 0}$
The sum on $k$ just equals the constant times the number of terms. Assuming the integer square root, $$S= \sum\limits_{i=1}^N \sum\limits_{j=1}^i \sum\limits_{k=1}^{i\lfloor\sqrt{j}\rfloor}C = \sum\limits_{i=1}^N \sum\limits_{j=1}^i Ci\lfloor\sqrt{j}\rfloor = C\sum\limits_{i=1}^N i\sum\limits_{j=1}^i\lfloor\sqrt{j}\rfloor.$$
The sum on $j$ is uneasy and can be addressed by decomposing into sequences that yield the same square root. For convenience, we will approximate it as if it was an integral: $$S\approx C\sum\limits_{i=1}^N i\left(\frac23i^{3/2}\right).$$ And finally for the sum on $i$, $$S\approx C\frac4{21}N^{7/2}.$$
Summing integer square roots
When $l^2\le j<(l+1)^2$, we have $\lfloor\sqrt j\rfloor=l$. There are exactly $(l+1)^2-l^2=2l+1$ values of $j$ that achieve this. The range $(0,j)$ can be decomposed in the full subranges $(l^2,(l+1)^2-1)$, for $l$ in $(0,\lfloor\sqrt j\rfloor-1)$, and the final incomplete interval $(\lfloor\sqrt j\rfloor^2,j)$.
Hence,
$$\sum\limits_{j=1}^i\lfloor\sqrt j\rfloor=\sum_{l=0}^{\lfloor\sqrt i\rfloor-1}(2l+1)l+(i-\lfloor\sqrt i\rfloor^2+1)\lfloor\sqrt i\rfloor.$$
The sum on $l$ is derived from Faulhaber's formula as
$$2\sum_{l=0}^{\lfloor\sqrt i\rfloor-1}l^2+\sum_{l=0}^{\lfloor\sqrt i\rfloor-1}l=2\frac{2\lfloor\sqrt i\rfloor^3-3\lfloor\sqrt i\rfloor^2+\lfloor\sqrt i\rfloor}6+\frac{\lfloor\sqrt i\rfloor(\lfloor\sqrt i\rfloor-1)}2$$ This confirms the $\frac23i^{3/2}$ behavior.
