3. Calculate $\left[\sqrt{2}+\sqrt[3]{\frac{3}{2}}+\sqrt[4]{\frac{4}{3}}+\ldots+\sqrt[2009]{\frac{2009}{2008}}\right]$ (here $[x]$ is the integer part of the number $x$, i.e., the greatest integer not exceeding the number $x$).
2.I.3. Since each of the numbers is greater than 1, their sum is greater than 2008. Let's prove that this sum is less than 2009. By the inequality between the geometric mean and the arithmetic mean, we have

$$
\sqrt[n+1]{\frac{n+1}{n}}=\sqrt[n+1]{\left(1+\frac{1}{n}\right) \cdot 1^{n}}<\frac{n+1+\frac{1}{n}}{n+1}=1+\frac{1}{n(n+1)}=1+\frac{1}{n}-\frac{1}{n+1}
$$

## Therefore

$$
\begin{aligned}
& \sqrt{2}+\sqrt[3]{\frac{3}{2}}+\sqrt[3]{\frac{4}{3}}+\ldots+\sqrt[2009]{\frac{2009}{2008}}< \\
&<2008+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\ldots+\frac{1}{2008}-\frac{1}{2009}= \\
&=2009-\frac{1}{2009}<2009
\end{aligned}
$$

Therefore, the integer part of the given sum is 2008.

Remark. The inequality $\sqrt[n+1]{1+\frac{1}{n}}<1+\frac{1}{n(n+1)}$ also follows from an analogue of Bernoulli's inequality for exponents less than 1.