amparvardi wrote: Find the integer represented by $\left[ \sum_{n=1}^{10^9} n^{-2/3} \right] $. Here $[x]$ denotes the greatest integer less than or equal to $x.$ $\int_1^{10^9}x^{-\frac 23}dx$ $>\sum_{n=2}^{10^9}n^{-\frac 23}$ $>\int_2^{10^9+1}x^{-\frac 23}dx$ $2997=3\sqrt[3]{10^9}-3\sqrt[3]{1}$ $>\sum_{n=2}^{10^9}n^{-\frac 23}$ $>3\sqrt[3]{10^9+1}-3\sqrt[3]{2}$ $>3000-4=2996$ $2998>\sum_{n=1}^{10^9}n^{-\frac 23}$ $>2997$ Hence the result : $\boxed{\left\lfloor\sum_{n=1}^{10^9}n^{-\frac 23}\right\rfloor=2997}$

A Non-Calculus Solution: Let $S:=\sum_{n=1}^{N^3}\frac{1}{n^{2/3}}$, where $N>1$ is an integer. Observe that \[\sum_{n=1}^{N^3}\frac{(n+1)^{1/3}-n^{1/3}}{(n+1)-n}<\frac{S}{3}<\frac{1}{3}+\sum_{n=2}^{N^3} \frac{n^{1/3}-(n-1)^{1/3}}{n-(n-1)}\,.\] Therefore, \[3(N-1)<3\left(\sqrt[3]{N^3+1}-1\right) < S < 1+3\left(\sqrt[3]{N^3}-1\right)\,.\] That is, $3N-3<S<3N-2$, so $\lfloor{S}\rfloor = 3N-3$. Plug in $N:=10^3$ to arrive at the final answer.