# Let $2, 1+\frac{1}{2}, 3, 1+\frac{1}{3}, 4, 1+\frac {1}{4},\dots$ be a sequence. Does $a_n$ converge/diverge? Is there a $\sup$ or $\inf$?

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?

1. $$a_n$$ coverges to a finitie limit or diverges to infinity.
2. $$\limsup \limits_{n \to \infty} (a_n) = \sup \{a_n| n \in \Bbb N\}$$
3. $$\liminf \limits_{n \to \infty} (a_n) = \inf\{a_n| n \in \Bbb N\}$$
4. The sequence $$a_n$$ has at least $$3$$ subsequential limits.
5. 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!

• If there is no formula for the general term then nothing can be said. I personally don't see an obvious pattern.
– 温泽海
May 22, 2022 at 14:29
• @温泽海 I do see an obvious pattern: Look at every other term. Now, is this pattern that I see the true pattern that the problem author intended? That is indeed impossible to tell without a more general description of the terms. May 22, 2022 at 14:47
• @温泽海 I thought that the sequence is supposed to continue the same as they did in the first elements and this way the even indexes are zero , I tried looking for a formula but did not work well May 22, 2022 at 14:49

1. This is not correct because the sequence does not converge, because as you pointed out $$\lim\limits_{n\to\infty} a_{2n-1} = \infty$$. The sequence also doesn't diverge to $$\infty$$ because it keeps oscillating between a large integer and a number close to $$1$$.
2. You actually could say this is true, depending on how your textbook/teacher defined the supremum $$\limsup\limits$$. The supremum could be said to be $$\infty$$, meaning there's no upper bound to the sequence, and because we just found a subsequence with limit $$\infty$$ we know that $$\limsup\limits_{n\to\infty} a_n = \infty$$.
3. This is definitely correct. You found a subsequence with limit $$1$$, and that's the infimum for the sequence, so $$\liminf\limits_{n\to\infty} a_n = 1$$.
4. I'm not sure if just because we "covered" the sequence with those two subsequences that that means there's no other subsequential limits, but it seems like the only obvious limits are $$1$$ and $$\infty$$. You could probably say that a subsequence with an infinite number of the odd numbered elements can only have $$\infty$$ as a limit. A subsequence with a finite number of the odd numbered elements would eventually turn into a sequence of just the even numbered elements, which go to $$1$$.
• Actually, $\liminf_{n\to\infty}a_n=1$. May 22, 2022 at 17:54