# Task 4. (12 points)

Which of the numbers is greater $\frac{1}{2016}\left(1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{2016}\right)$ or $\frac{1}{2017}\left(1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{2017}\right)$?
Solution.

Let

$$
x=\frac{1}{2016}\left(1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{2016}\right), y=\frac{1}{2017}\left(1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{2017}\right), a=\left(1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{2016}\right)
$$

Then $x=\frac{a}{2016}, y=\frac{1}{2017}\left(a+\frac{1}{2017}\right), x-y=\frac{a}{2016}-\frac{a}{2017}-\frac{1}{2017^{2}}=\frac{1}{2017}\left(\frac{a}{2016}-\frac{1}{2017}\right)>0$,

since $a>1$.

Answer: the first number is greater.

| Criterion | Evaluation | Points |
| :---: | :---: | :---: |
| Complete solution. | + | 12 |
| All logical steps of the solution are presented. Due to a computational error, an incorrect answer is obtained. | $\pm$ | 9 |
| The relationship between the two compared numbers is shown. There are errors in algebraic transformations when comparing numbers. The answer is provided, possibly incorrect. | $+/ 2$ | 6 |
| The relationship between the two compared numbers is shown. There is no complete justification for the correct answer. |  |  |
| The method of enumerating possible price changes was used in solving the problem. More than two options were not presented. The answer is correct. | $\mp$ | 2 |
| The relationship between the two compared numbers is shown. There is no answer. |  |  |
| The solution does not meet any of the criteria described above. | $-/ 0$ | 0 |
| Maximum score |  | 12 |