1. Determine the integer part of the number $x=\frac{1}{2^{2}}+\frac{1}{3^{2}}+\frac{1}{4^{2}}+\ldots . .+\frac{1}{2013^{2}}$.

Prof. Andronic Aurica Mihaela
Solution: It is observed that each term of the sum satisfies: $\frac{1}{2 \cdot 3}<\frac{1}{2^{2}}<\frac{1}{1 \cdot 2}$;

$\frac{1}{3 \cdot 4}<\frac{1}{3^{2}}<\frac{1}{2 \cdot 3} ; \frac{1}{4 \cdot 5}<\frac{1}{4^{2}}<\frac{1}{3 \cdot 4} ; \ldots \ldots \ldots \ldots . . \frac{1}{2013 \cdot 2014}<\frac{1}{2013^{2}}<\frac{1}{2012 \cdot 2013}$. Adding the inequalities, we obtain:

$\frac{1}{2 \cdot 3}+\frac{1}{3 \cdot 4}+\frac{1}{4 \cdot 5}+\ldots . .+\frac{1}{2013 \cdot 2014}<\frac{1}{2^{2}}+\frac{1}{3^{2}}+\frac{1}{4^{2}}+\ldots .+\frac{1}{2013^{2}}<\frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\frac{1}{3 \cdot 4}+\ldots .+\frac{1}{2012 \cdot 2013}$ Thus, $\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\ldots . .+\frac{1}{2013}-\frac{1}{2014}<\frac{1}{2^{2}}+\frac{1}{3^{2}}+\frac{1}{4^{2}}+\ldots .+\frac{1}{2013^{2}}<1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\ldots .+\frac{1}{2012}-\frac{1}{2013}$. From which it follows: $\frac{1}{2}-\frac{1}{2014}<\frac{1}{2^{2}}+\frac{1}{3^{2}}+\frac{1}{4^{2}}+\ldots .+\frac{1}{2013^{2}}<1-\frac{1}{2013} \Rightarrow \frac{1006}{2014}<x<1-\frac{1}{2013} \Rightarrow 0 \leq x<1 \Rightarrow[x]=0$.

## Scoring Guide:

| $\frac{1}{2 \cdot 3}<\frac{1}{2^{2}}<\frac{1}{1 \cdot 2} ; \frac{1}{3 \cdot 4}<\frac{1}{3^{2}}<\frac{1}{2 \cdot 3} ; \frac{1}{4 \cdot 5}<\frac{1}{4^{2}}<\frac{1}{3 \cdot 4} ; \ldots ; \frac{1}{2013 \cdot 2014}<\frac{1}{2013^{2}}<\frac{1}{2012 \cdot 2013}$ | $\mathbf{2} \mathbf{p}$ |
| :--- | :--- |
| $\frac{1}{2 \cdot 3}+\frac{1}{3 \cdot 4}+\frac{1}{4 \cdot 5}+\ldots \ldots+\frac{1}{2013 \cdot 2014}<\frac{1}{2^{2}}+\frac{1}{3^{2}}+\frac{1}{4^{2}}+\ldots .+\frac{1}{2013^{2}}<\frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\frac{1}{3 \cdot 4}+\ldots .+\frac{1}{2012 \cdot 2013}$ | $\mathbf{1} \mathbf{p}$ |
| $\frac{1}{2}-\frac{1}{2014}<\frac{1}{2^{2}}+\frac{1}{3^{2}}+\frac{1}{4^{2}}+\ldots .+\frac{1}{2013^{2}}<1-\frac{1}{2013}$ | $\mathbf{2} \mathbf{p}$ |
| $\frac{1006}{2014}<x<1-\frac{1}{2013} \Rightarrow 0 \leq x<1$ | $\mathbf{1} \mathbf{p}$ |
| $[x]=0$ | $\mathbf{1} \mathbf{p}$ |