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Given the integral identity \begin{align} \int_{0}^{t} \cdots \int_{0}^{t - t_{1} - \dots - t_{n -1}} 1 \ dt_1 \cdots dt_n = \frac{t^{n}}{n!}, \end{align} I believe it is true that \begin{align} \sum_{i_{1} = 0}^{\lfloor t \rfloor} \cdots \sum_{i_{n} = 0}^{\lfloor t - i_1 - \cdots - i_{n-1} \rfloor} 1 = \frac{t^{n}}{n!} + O(t^{n-1}), \end{align} where $\lfloor \cdot \rfloor$ is the floor function. However, I'm having difficulty in justifying the exponent $n - 1$. Is it indeed the case? Any help would certainly be appreciated.

Thanks!

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up vote 9 down vote accepted

The sum is equal to $\binom{\lfloor t \rfloor + n}{n}$ (choose the distinct numbers $i_1+1, i_1+i_2+2, \ldots , i_1 + \ldots + i_n + n$ in $[ 1 , \lfloor t \rfloor + n ]$), so it is a polynomial in $\lfloor t \rfloor$ of degree $n$ and leading coefficient $1/n!$, and the asymptotic behavior you guessed is correct.

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