3. (USA) 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 $x$ (e.g. $[\sqrt{2}]=1$ ).

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3. From $\left((k+1)^{2 / 3}+(k+1)^{1 / 3} k^{1 / 3}+k^{2 / 3}\right)\left((k+1)^{1 / 3}-k^{1 / 3}\right)=1$ and $3 k^{2 / 3}<(k+1)^{2 / 3}+(k+1)^{1 / 3} k^{1 / 3}+k^{2 / 3}<3(k+1)^{2 / 3}$ we obtain 
$$ 3\left((k+1)^{1 / 3}-k^{1 / 3}\right)<k^{-2 / 3}<3\left(k^{1 / 3}-(k-1)^{1 / 3}\right) . $$ 
Summing from 1 to $n$ we get 
$$ 1+3\left((n+1)^{1 / 3}-2^{1 / 3}\right)<\sum_{k=1}^{n} k^{-2 / 3}<1+3\left(n^{1 / 3}-1\right) $$ 
In particular, for $n=10^{9}$ this inequality gives 
$$ 2997<1+3\left(\left(10^{9}+1\right)^{1 / 3}-2^{1 / 3}\right)<\sum_{k=1}^{10^{9}} k^{-2 / 3}<2998 $$ 
Therefore $\left[\sum_{k=1}^{10^{9}} k^{-2 / 3}\right]=2997$.