How do i evaluate this sum?

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Let $[x]$ be the nearest integer to $x$. (for $x=n+\frac{1}{2}, n \in N$, let $[x]=n$).

Find the value of $$\displaystyle\sum_{m=1}^{\infty} [\sqrt m]^{-3}$$

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Hint: note that $[m]=1$ twice, it equals $2$ four times (from $3$ through $6$). How many times does it equal $k$?