limit/limsup/liminf of a sequence (by an example)

Let us check for limit/limsup/liminf of the sequence \(\frac{n}{10^{\lceil \log_{10}n \rceil}}\), where the notation \(\lceil x \rceil\) means the smallest integer greater than or equal to \(x\).

For example, \(\lceil 0.1 \rceil=1, \lceil 0.9 \rceil=1, \lceil -1.2 \rceil=-1, \lceil -2.5 \rceil=-2\)

To compute limit (to have a hope of computing), we need to know it converge (which we can check by checking it is Cauchy sequence).

To compute limsup/liminf, we do not need it to be Cauchy or we do not need to know if it is Cauchy. All that matters is if it is bounded.

Is it bounded?

Bounded below, for sure, as every element is non-negative.

Is it bounded above?

Let's check.

If it is not bounded above, then, for 11924, there exists an \(n\in \mathbb{N}\) with the property that,

\[\frac{n}{10^{\lceil \log_{10}n \rceil}}>11924\]

which is equivalent to saying (or at least imply that)

\[\log_{10}(n)-\log_{10}(10^{\lceil \log_{10}n \rceil})>\log(11924)\]

which is equivalent to saying (or at least imply that)

\[\log_{10}(n)-\lceil \log_{10}n \rceil>\log(11924)\]

which is just not true.

The difference between \(x\) and \(\lceil x \rceil\) is less than \(1\). To be precise, \(\lceil x \rceil-x<1\).

So, we can not have \(\log_{10}(n)-\lceil \log_{10}n \rceil>\log(11924)\) as \(\log_{10}(11924)>0\) where as \(\log_{10}(n)-\lceil \log_{10}n \rceil<0\).

Or for that matter, we can not have \(\log_{10}(n)-\lceil \log_{10}n \rceil>\log_{10}(1924)\) as \(\log(1924)>0\) where as \(\log_{10}(n)-\lceil \log_{10}n \rceil<0\).

We can not even have \(\log_{10}(n)-\lceil \log_{10}n \rceil>\log_{10}(10)\) as \(\log(1924)>0\) where as \(\log_{10}(n)-\lceil \log_{10}n \rceil<0\).

How about \(\log_{10}(n)-\lceil \log_{10}n \rceil>\log_{10}(1)\)??

That is also not true, as \(\log_{10}(1)\geq 0\) where as \(\log_{10}(n)-\lceil \log_{10}n \rceil<0\).

So, there is no \(n\in \mathbb{N}\) with the property that \(\log_{10}(n)-\lceil \log_{10}n \rceil>\log_{10}(1)\) which means, for every \(n\in \mathbb{N}\), we have \(\log_{10}(n)-\lceil \log_{10}n \rceil<\log_{10}(1)\)

which is equivalent to saying (or at least imply that)

\[\log_{10}(n)<\log_{10}(1)+\lceil \log_{10}n \rceil=\log_{10}(1)+\log_{10}(10^{\lceil \log_{10})n \rceil}=\log_{10}(10^{\lceil \log_{10})n \rceil}\]

which is equivalent to saying (or at least imply that),

\[n<10^{\lceil \log_{10}n \rceil}\] for all \(n\in \mathbb{N}\),

which is equivalent to saying (or at least imply that),

\[\frac{n}{10^{\lceil \log_{10}n \rceil}}<1\] for all \(n\in \mathbb{N}\),

Thus the sequence \((a_n)\) is bounded above by \(1\).

This gives us idea about what to expect as \(\lim(a_n)\) (if someone who we believe tells us that \((a_n)\) converges, but cant say anything more than that).

As \((a_n)\) is bounded above by \(1\), the limit (if it exists) can be almost \(1\). If the limit does not exists, the \(\limsup (a_n)\) is also almost \(1\).

There is a thing about dealing with log base 10. It behaves in a certain way when \(n\) is a power of \(10\).

\[\log_{10}10=1, \log_{10}(10^2)=2, \log_{10}(10^3)=3,\cdots\],

which assures us that,

\[\lceil \log_{10}10 \rceil=1, \lceil \log_{10}(10^2) \rceil=2, \lceil \log_{10}(10^3) \rceil=3,\cdots\],

which imply that

\[10^{\lceil \log_{10}10 \rceil}=10^1, 10^{\lceil \log_{10}(10^2) \rceil}=10^2, 10^{\lceil \log_{10}(10^3) \rceil}=10^3,\cdots\],

or, more generally, \(10^{\lceil \log_{10}(10^k) \rceil}=10^k\) for all \(k\in \mathbb{N}\), which imply that,

\[\frac{10^k}{10^{\lceil \log_{10}(10^k) \rceil}}=\frac{10^k}{10^k}=1\] for all \(k\in \mathbb{N}\)

Great. It is not just that \(1\) is an upper bound for the sequence, but, it actually appears at multiple places. We actually have a subsequence of \((a_n)\) that is just the constant sequence \(1\)

Boom!!! \(1\) is the limsup of \((a_n)\).

If \((a_n)\) is bounded above by \(L\in \mathbb{R}\) and we are able to get a subsequence \((a_{n_k})\) that converges to \(L\), then, \(\limsup(a_n)=L\).

This does not say anything about the non-convergence of the sequence \((a_n)\).

It may happen that the sequence converges to \(1\).

It would not happen if we are able to find another subsequence that converges to anything else but \(1\).

We know the behavior of the subseqeunce \((a_{10^k})\). Keeping it as the reference subsequence, let us check the subsequences to the immediate left and immediate right of \((a_{10^k}\).

Consider the subsequence \(a_{10^k-1}\). We have

\[a_{10^k-1}=\frac{10^k-1}{10^{\lceil \log_{10}(10^k-1) \rceil}}\]

As \(10^k-1<10^k\), we would have \(\log_{10}(10^k-1)<\log_{10}(10^k)=k\).

As there is not much of a gap between \(10^k-1\) and \(10^k\), we can safely believe (you should prove) that, \(\lceil \log_{10}(10^k-1) \rceil=k\). Thus,

\[a_{10^k-1}=\frac{10^k-1}{10^{\lceil \log_{10}(10^k-1) \rceil}}=\frac{10^k-1}{10^k}=1-\frac{1}{10^k}\]

It is unfortunate that this subsequence also converges to \(1\). So, we can not conclude if \((a_n)\) converges or not.

Let's continue with the subsequence that is immediate right to \((a_{10^k})\), the subsequence \(a_{10^k+1}\). We have

\[a_{10^k+1}=\frac{10^k+1}{10^{\lceil \log_{10}(10^k+1) \rceil}}\].

Unlike \(10^k-1\), the component \(10^k+1\) behaves in a slightly different way.

As \(10^k<10^k+1\), we have \(\log_{10}(10^k)<\log_{10}(10^{k}+1)\).

Which implies, \(\lceil \log_{10}(10^k) \rceil<\lceil \log_{10}(10^{k}+1) \rceil\); that is, \(k<\lceil \log_{10}(10^{k}+1) \rceil\), which is same as saying \(\lceil \log_{10}(10^{k}+1) \rceil=k+1\). So,

\[a_{10^k+1}=\frac{10^k+1}{10^{\lceil \log_{10}(10^k+1) \rceil}}=\frac{10^k+1}{10^{k+1}}=\frac{10^k+1}{10^{k+1}}=\frac{1}{10}+\frac{1}{10^k}\].

This subsequence converges to \(\frac{1}{10}\) as the sequence \(\frac{1}{10^k}\) converges to \(0\). It gives comfort when we realize there is a subsequence that does not converge to \(1\). This assures the sequence \((a_n)\) is not convergent. We were able to find \(\limsup(a_n)\) just from one subsequence. That is because an upper bound is matching with the limit of subsequence.

We can not conclude \(\liminf\) is \(\frac{1}{10}\) as we may not be able to assure there is no subsequence that converges to an element less than \(\frac{1}{10}\).

We can assure if somehow we can prove that \(\frac{1}{10}\) is a lower bound for the sequence.

If \((a_n)\) is bounded below by \(L\in \mathbb{R}\) and we are able to get a subsequence \((a_{n_k})\) that converges to \(L\), then, \(\liminf(a_n)=L\).

It is easy to check if \(\frac{1}{10}\) is a lower bound or nor. All we need to see is if, for all \(n\in \mathbb{N}\), we have

\[\frac{1}{10}\leq \frac{n}{10^{\lceil \log_{10}n \rceil}}\]

which is same as saying

\[10^{\lceil \log_{10}n \rceil}<10 n\],

which is same as saying

\[\log_{10}(10^{\lceil \log_{10}n \rceil})<\log_{10}(10 n)=\log_{10}10+\log_{10}n\]

which is same thing as saying

\(\lceil \log_{10}n \rceil< 1+\log_{10}n\),

which is as true as the statement "sun rises in the east".

So, the sequence \((a_n)\) is bounded below by \(\frac{1}{10}\) and we have a subsequence that converges to \(\frac{1}{10}\).

So, \(\liminf a_n=\frac{1}{10}\).