Find the Supremum, Infimum, lim sup, and lim inf of 2 − 1/n

Assume $n$ ranges over the positive integers, so the sequence starts at $n=1$. The first terms are $1,\; 3/2,\; 5/3,\; 7/4,\ldots$. They increase toward $2$, but none equals $2$.

Therefore, the range has infimum $1$ and supremum $2$. The infimum is attained as the minimum; the supremum is not attained, so there is no maximum. The sequence converges to $2$, which means its limit superior and limit inferior are both $2$.

Find the infimum and supremum

For every positive integer $n$, we have $1/n>0$ and $1/n\leq 1$. Subtracting from $2$ gives

$$1\leq 2-\frac{1}{n}<2.$$

The lower bound is attained at $n=1$, since $a_1=2-1=1$. Thus $1$ is the minimum of the sequence values and, consequently, their infimum:

$$\inf\{a_n:n\in\mathbb{N},\ n\geq 1\}=1.$$

The same inequality shows that $2$ is an upper bound. To see that it is the least upper bound, take any $\varepsilon>0$. Choose a positive integer $n>1/\varepsilon$. Then

$$2-\varepsilon<2-\frac{1}{n}<2.$$

So terms get arbitrarily close to $2$ from below. No number smaller than $2$ can be an upper bound, and therefore

$$\sup\{a_n:n\in\mathbb{N},\ n\geq 1\}=2.$$

Since $a_n<2$ for every $n$, this supremum is not attained: the sequence has no maximum.

Compute lim sup and lim inf

The limit superior and limit inferior describe the limiting behavior of the tails of a sequence. For a bounded sequence, they are defined by

$$\limsup_{n\to\infty}a_n=\lim_{N\to\infty}\sup_{n\geq N}a_n, \qquad \liminf_{n\to\infty}a_n=\lim_{N\to\infty}\inf_{n\geq N}a_n.$$

Here the sequence is increasing, since

$$a_{n+1}-a_n=\left(2-\frac{1}{n+1}\right)-\left(2-\frac{1}{n}\right)=\frac{1}{n(n+1)}>0.$$

Consequently, in the tail beginning at $N$, the smallest term is $a_N$, and the tail supremum is $2$. The latter is a supremum, not a maximum: all terms in every tail remain strictly below $2$. Hence

$$\inf_{n\geq N}a_n=a_N=2-\frac{1}{N}, \qquad \sup_{n\geq N}a_n=2.$$

As $N$ tends to infinity, $a_N$ tends to $2$, while the tail suprema are already equal to $2$. Thus

$$\boxed{\liminf_{n\to\infty}a_n=2,\qquad \limsup_{n\to\infty}a_n=2.}$$

This also follows from the usual limit: $a_n=2-1/n\to2$. Whenever a sequence converges to a finite limit, its lim inf and lim sup equal that limit.

A finite-term check

The first five terms are

$$1,\quad \frac{3}{2},\quad \frac{5}{3},\quad \frac{7}{4},\quad \frac{9}{5}.$$

For these five terms alone, the minimum is $1$ and the maximum is $9/5$. For the whole infinite sequence, however, terms continue beyond $9/5$ and approach $2$. This illustrates why the maximum of a finite initial segment is not necessarily the supremum of the complete sequence.

Common mistakes

• Calling $2$ the maximum: A maximum must be a term of the sequence. Here $2-1/n<2$ for every finite positive $n$, so $2$ is only the supremum.
• Confusing the first term with the limit: The first term is $1$; the terms approach $2$. The infimum of the range is not the sequence’s limit.
• Using only the limit to find the infimum: Knowing the limit is $2$ does not by itself identify the infimum. Check all terms; here they are at least $1$, and $a_1=1$.
• Ignoring the indexing convention: Some texts use $\mathbb{N}$ for positive integers, while others include zero. Since $2-1/n$ is undefined at $n=0$, this sequence must be indexed by positive integers.

Quantity	Value	Attained?
Infimum	$1$	Yes; it is the minimum at $n=1$
Supremum	$2$	No; it is not a term
Limit inferior	$2$	Not an extremum of the original range
Limit superior	$2$	Not an extremum of the original range