10. What is greater: $\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+\ldots+\frac{1}{2010}-\frac{1}{2011}$ or $\frac{1}{1006}+\frac{1}{1007}+\ldots+\frac{1}{2011}$?
Answer: the right number is greater.

Solution. Let's denote the sum of the form $1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{n}$ as $H_{n}$. Then the left part can be transformed as follows:

$$
\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+\ldots+\frac{1}{2010}-\frac{1}{2011}=2 \cdot\left(\frac{1}{2}+\frac{1}{4}+\ldots+\frac{1}{2010}\right)+1-H_{2011}=H_{1005}+1-H_{2011}
$$

and the right part can be transformed as follows:

$$
\frac{1}{1006}+\frac{1}{1007}+\ldots+\frac{1}{2011}=H_{2011}-H_{1005}
$$

It remains to compare $H_{1005}+1-H_{2011}$ and $H_{2011}-H_{1005}$. If $H_{1005}+1-H_{2011} \vee H_{2011}-H_{1005}$, then $H_{1005}+1 / 2 \vee H_{2011}$, that is, $\frac{1}{2} \vee \frac{1}{1006}+\frac{1}{1007}+\ldots+\frac{1}{2011}$. It is clear that

$$
\frac{1}{1006}+\frac{1}{1007}+\ldots+\frac{1}{2011}>\underbrace{\frac{1}{2010}+\frac{1}{2010}+\ldots+\frac{1}{2010}}_{1005 \text { terms }}+\frac{1}{2011}=\frac{1}{2}+\frac{1}{2011}>\frac{1}{2}
$$