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I have a slight issue with the mark scheme's approach to the following problem (I used a different definition, but nevertheless got the same answer).

Let ${(a_n)}_{n\in \mathbb{N}}$ be a sequence defined by:

$a_{2n}= 1+ \frac{1}{1+n}, a_{2n+1}= ln(1+ \frac{1}{1+n})$.

I have to find lim sup and lim inf.

Here is what I've done:

For lim sup, for instance:

I have to find $\lim_{n \to \infty} b_n$ where $b_n = \sup\{a_k : k \geq n \}$.

I've defined $b_n$ as follows:

If $n$ is even, then: $b_n = 1+ \frac{1}{1+n}$

(As $b_n$ is decreasing and $x> lnx$)

If $n$ is odd, then: $b_n = 1+ \frac{1}{2+n}$

(I kind of guessed here, as I thought that every even term which follows an odd term will be greater than that odd term due to the log property above. So I've just incremented $n$ by 1 to make it even).

This is where the markscheme disagrees, as $b_n$ is defined differently for both scenarios:

If $n$ is even, then: $b_n = 1+ \frac{1}{1+n/2}$

If $n$ is odd, then: $b_n = 1+ \frac{1}{1 + (n+1)/2}$

The mark schemes approach makes little sense to me as we are only supposed to consider the values of the sequence which have indices greater or equal to n (and take their supremum), whereas in this case, $n$ is divided by 2 which makes it seem that we are considering terms that have indices smaller than $n$. What is going on here? What is wrong with my approach?

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    If $n$ is even, say $n = 2k$, then $b_n = a_n = a_{2k} = 1 + \frac{1}{k}$, not $1 + \frac{1}{n}$. And if $n$ is odd, $n = 2k+1$, then $b_n = a_{n+1} = a_{2k+2} = 1 + \frac{1}{1 + (k+1)}$. – 2017-01-08
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    @DanielFischer thanks for your response. Now I understand that I've simply confused the indices! – 2017-01-09

2 Answers 2

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Recall that if for some $p$, the subsequences $(a_{pn + r})$, for $0\le r\le p-1$, are all convergent, then the set of cluster points of $(a_n)$ is the set of all limits of these subsequences. Looking at our sequence, $a_{2n} \to 1$ and $a_{2n+1} \to 0$, hence the set of cluster points of $(a_n)$ is $\{0,1\}$, and $\liminf a_n = \inf \{0,1\} = 0$, and $\limsup a_n = \sup \{0,1\} = 1$.

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$\lim_{n\to \infty}a_{2n}=1$ and $\lim_{m\to \infty}a_{2n+1}=0.$

So for any $r\in (0,1/2)$ there exists $m$ such that $$\{a_{2n}:2n>m\}\subset (-r+1,r+1) \;\land \; \{a_{2n+1}:2n+1>m\}\subset (-r,r).$$ So $\lim \sup a_n\in [-r+1,r+1]$ and $\lim \inf a_n \in [-r,r].$ Since this holds for every $r\in (0,1/2)$ we have $\lim \inf a_n=0$ and $\lim \sup a_n=1.$

In the style of the 17th and 18th centuries: "When $n$ is infinitely large then $a_{2n}$ is infinitely close to $1$ and $a_{2n+1}$ is infinitely close to $0.$"