Let $2,1+\frac{1}{2},3,1+\frac{1}{3},4,1+\frac {1}{4},...$ be a sequence then which of the statements is true?
- $a_n$ coverges to a finitie limit or diverges to infinity.
- $\limsup \limits_{n \to \infty} (a_n) = \sup \{a_n| n \in \Bbb N\}$
- $\liminf \limits_{n \to \infty} (a_n) = \inf\{a_n| n \in \Bbb N\}$
- The sequence $a_n$ has at least $3$ subsequential limits.
- None of the above.
For the first statement I could not actually find a formula, but it is obvious that if we look at the subsequence in the even indexes we get $\lim \limits_{n \to \infty}a_{2n} =1 $ and in odd $\lim \limits_{n \to \infty}a_{2n-1} = \infty $ so the limit does not exist therefore the statement is not correct.
For the second statement I assumed it was bounded from above, therefore there exists an $M>1$ such that for all $n$ we get $a_n \leq M$ , let $L$ be a sub sequential limit of $a_n$ so there exists a subsequence $a_{n_k}$ of $a_n$ such that $\lim \limits_{n \to \infty}a_{{n_k}} =L$ for all $n$ we have $a_n \leq M$ $\implies$ for all $k$ we have $a_{n_k} \leq M$ so we get $L = \lim \limits_{n \to \infty}a_{n_k} \leq M$ meaning all of the sub sequential limits of $a_n$ are less or equal to $M$ because one subsequential limit is $1$ and the other is infinity so it is not bounded from above and there is not $\sup$.
I thought the third statement is correct because the sequence is bounded from below by $1$ so it is the minimum, but according to the answers in the book it is not the correct answer and I could not figure out why.
For the fourth statement, as we found in the first part there are two different limits that one is even and the other is odd indexes that cover the sequence so there are only two limits so it is not true.
The right answer according to the book is the fifth statement but why isn't the third statement correct?
Thank you for the amazing help and tips!