Problems from Arbitrary union and intersection

Development:

1. As the lower endpoint of all the intervals is the same, to find the intersection, we have to find the infimum of the upper endpoints.

   To find it, we are going to study the sequence of endpoints:

   $$\Big\{\dfrac{2n^2}{3n+1}\Big\}_{n\in\mathbb{N}}$$, it is a divergent sequence (since the degree of the numerator is bigger than the degree of the denominator), and it is also an increasing sequence, and therefore, the minimum is the first element of the sequence: $$$a_2=\dfrac{2\cdot 2^2}{3\cdot 2 + 1}=\dfrac{8}{7}$$$ And so, we have the intersection: $$$\bigcap_{n\geq 2}\Big(1,\dfrac{2n^2}{3n+1}\Big)=\Big(1,\dfrac{8}{7}\Big)$$$

   Notice that if we should add the first member of the sequence, since $$$a_1=\dfrac{2\cdot 1^2}{3\cdot1+1}=\dfrac{1}{2}$$$ We would have to add the interval $$\Big(\dfrac{1}{2},1\Big)$$, and then the intersection would be empty: $$$\bigcap_{n\geq 1}\Big(1,\dfrac{2n^2}{3n+1}\Big)=\emptyset$$$

2. To study this union, we are going to study the behavior of the sequences of endpoints.

   Starting with the lower endpoint:

   $$\Big\{-\dfrac{5n-1}{n+52}\Big\}_{n\in\mathbb{N}}$$ it is a sequence that is convergent to $$-5$$ and decreasing:

   $$$-\dfrac{5n-1}{n+52} > -\dfrac{5(n+1)-1}{n+1+52}\dfrac{5n-1}{n+52} < \dfrac{5n+4}{n+53} $$$ $$$ (5n-1)(n+53) < (5n+4)(n+52)$$$ $$$ 5n^2+265n-n-53 < 5n^2+260n+4n+208$$$ $$$ 5n^2+264n - 53 < 5n^2+260n+4n+208 $$$ $$$ -53 < 208 $$$

   And so, the lower endpoint of the union is the limit of the sequence, which is $$-5$$.

   Let's study now the upper limit. The endpoints sequence is:

   $$\Big\{\dfrac{7n^2}{8+n}\Big\}_{n\in\mathbb{N}}.$$ As the degree of the numerator is bigger than the degree of the denominator, we know that it is an increasing sequence and divergent to $$+\infty$$, this being the supremus of this sequence.

   That is, we have: $$$\bigcup_{n\geq 1}\Big[-\dfrac{5n-1}{n+52},\dfrac{7n^2}{8-n}\Big]=[-5,+\infty)$$$

Solution:

1. $$\bigcap_{n\geq 2}\Big(1,\dfrac{2n^2}{3n+1}\Big)=\Big(1,\dfrac{8}{7}\Big)$$
2. $$\bigcup_{n\geq 1}\Big[-\dfrac{5n-1}{n+52},\dfrac{7n^2}{8-n}\Big]=[-5,+\infty)$$

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