## Task 1 - 201221

For each $n=1,2,3, \ldots$ let

$$
a_{n}=\frac{1}{n^{2}} \sum_{k=1}^{n} k
$$

Furthermore, let $I_{1}, I_{2}, I_{3}$, and $I_{4}$ be the closed intervals

$$
I_{1}=[1 ; 2], \quad I_{2}=[0.53 ; 0.531], \quad I_{3}=[0.509 ; 0.51], \quad I_{4}=[0.4 ; 0.5]
$$

Investigate for each of these intervals whether there are terms of the sequence $\left\{a_{n}\right\}$ lying in them.

If this is the case, determine the indices $n$ of all terms $a_{n}$ in the respective interval.
For each $n=1,2,3, \ldots$, we have

$$
\left.\left.\sum_{k=1}^{n} k=\frac{1}{2} n(n+1), \quad \text { so } \quad a_{n}=\frac{1}{2 n^{2}}\right) n^{2}+n\right)=\frac{1}{2}+\frac{1}{2 n}
$$

Thus, $a_{n} \in I_{1}, a_{n} \in I_{2}, a_{n} \in I_{3}$ or $a_{n} \in I_{4}$ are respectively equivalent to the following inequalities:

- $a_{n} \in I_{1}$

$$
\begin{aligned}
& 1 \leq \frac{1}{2}+\frac{1}{2 n} \leq 2 \quad ; \quad \frac{1}{2} \leq \frac{1}{2 n} \leq \frac{3}{2} \\
& 2 \geq 2 n \geq \frac{2}{3} \quad ; \quad \frac{1}{3} \leq n \leq 1
\end{aligned}
$$

- $a_{n} \in I_{2}$

$$
\begin{aligned}
& 0.53 \leq \frac{1}{2}+\frac{1}{2 n} \leq 0.531 \quad ; \quad \frac{3}{100} \leq \frac{1}{2 n} \leq \frac{31}{1000} \\
& \frac{100}{3} \geq 2 n \geq \frac{1000}{31} \quad ; \quad \frac{500}{31} \leq n \leq \frac{50}{3}
\end{aligned}
$$

- $a_{n} \in I_{3}$

$$
\begin{aligned}
& 0.509 \leq \frac{1}{2}+\frac{1}{2 n} \leq 0.51 \quad ; \quad \frac{9}{100} \leq \frac{1}{2 n} \leq \frac{1}{100} \\
& \frac{1000}{9} \geq 2 n \geq 100 \quad ; \quad 50 \leq n \leq \frac{500}{9}
\end{aligned}
$$

- $a_{n} \in I_{4}$

$$
0.4 \leq \frac{1}{2}+\frac{1}{2 n} \leq 0.5 \quad ; \quad -0.1 \leq \frac{1}{2 n} \leq 0
$$

Since $160$ for all $n=1,2,3, \ldots$, it follows that:

$a_{n} \in I_{1}$ exactly for $n=1$,

$a_{n} \in I_{2}$ for no $n$,

$a_{n} \in I_{3}$ exactly for $n=50,51,52,53,54,55$,

$a_{n} \in I_{4}$ for no $n$.

## Adapted from $[5]$