Finding the Limit Point of the Sequence

The given sequence is <1, 2, 1/2, 3, 1/3, 4, 1/4,..... >.

Let's denote the sequence as $(a_n)_{n=1}^\infty$. The terms are:

• $a_1 = 1$
• $a_2 = 2$
• $a_3 = 1/2$
• $a_4 = 3$
• $a_5 = 1/3$
• $a_6 = 4$
• $a_7 = 1/4$
• and so on...

We can observe a pattern in the terms for $n \ge 2$. The terms with even indices seem to be increasing integers, and the terms with odd indices (starting from $a_3$) seem to be reciprocals of increasing integers.

Let's formalize this pattern:

• $a_1 = 1$
• For even indices $n \ge 2$, let $n=2k$ for some integer $k \ge 1$. The term is $a_{2k} = k+1$.
• For odd indices $n \ge 3$, let $n=2k+1$ for some integer $k \ge 1$. The term is $a_{2k+1} = 1/(k+1)$.

Let's check this pattern:

• $k=1$: $a_2 = 1+1 = 2$, $a_3 = 1/(1+1) = 1/2$. Matches the sequence.
• $k=2$: $a_4 = 2+1 = 3$, $a_5 = 1/(2+1) = 1/3$. Matches the sequence.
• $k=3$: $a_6 = 3+1 = 4$, $a_7 = 1/(3+1) = 1/4$. Matches the sequence.

Understanding Limit Points of a Sequence

A number $L$ is called a limit point (or accumulation point) of a sequence $(a_n)$ if every neighborhood of $L$ contains infinitely many terms of the sequence. Equivalently, $L$ is a limit point if there exists a subsequence of $(a_n)$ that converges to $L$.

Analyzing Subsequences

To find the limit points of the sequence, we can look at the behavior of its subsequences.

Subsequence of Even-Indexed Terms

Consider the subsequence formed by the even-indexed terms (starting from $a_2$). These terms are $a_2, a_4, a_6, a_8, ...$. Using our pattern, the terms are $k+1$ for $k=1, 2, 3, 4, ...$. The subsequence is <2, 3, 4, 5, ...>. Let's examine the limit of this subsequence:

$\lim_{k \to \infty} a_{2k} = \lim_{k \to \infty} (k+1)$

As $k$ becomes very large, $k+1$ also becomes very large. So, $\lim_{k \to \infty} (k+1) = \infty$. This subsequence diverges to infinity. Infinity is not considered a limit point in the set of real numbers.

Subsequence of Odd-Indexed Terms (from $a_3$)

Consider the subsequence formed by the odd-indexed terms starting from $a_3$. These terms are $a_3, a_5, a_7, a_9, ...$. Using our pattern, the terms are $1/(k+1)$ for $k=1, 2, 3, 4, ...$. The subsequence is <1/2, 1/3, 1/4, 1/5, ...>. Let's examine the limit of this subsequence:

$\lim_{k \to \infty} a_{2k+1} = \lim_{k \to \infty} \frac{1}{k+1}$

As $k$ becomes very large, $k+1$ becomes very large, and $1/(k+1)$ approaches 0. So, $\lim_{k \to \infty} \frac{1}{k+1} = 0$. This subsequence converges to 0. Therefore, 0 is a limit point of the sequence.

Examining the First Term ($a_1$)

The first term of the sequence is $a_1 = 1$. For 1 to be a limit point, it must appear infinitely often in the sequence, or there must be a subsequence converging to 1. The terms $a_n$ for $n \ge 2$ are either of the form $k+1$ (which are $2, 3, 4, ...$) or $1/(k+1)$ (which are $1/2, 1/3, 1/4, ...$). Neither of these sets of terms contains the value 1 (except possibly for specific initial values, but we are looking at infinitely many terms). Since the value 1 appears only once (as $a_1$), it cannot be a limit point based on this analysis.

Identifying All Limit Points

We have analyzed the main subsequences that cover almost all terms of the sequence (all terms except $a_1$). The subsequence of even terms diverges to infinity. The subsequence of odd terms (from $a_3$) converges to 0. The term $a_1=1$ does not form a convergent subsequence to 1. Thus, the only limit point of the sequence is 0.

Conclusion

Based on the analysis of the subsequences, the only limit point of the sequence <1, 2, 1/2, 3, 1/3,..... > is 0.

Comparing this with the given options:

• Option 1: 3
• Option 2: 0
• Option 3: 2
• Option 4: 1

The limit point we found is 0, which corresponds to Option 2.

Revision Table: Sequence Limit Point Analysis

Additional Information: Properties of Limit Points

Here are some important points about limit points of sequences in real analysis:

• A sequence can have none, one, or multiple limit points.
• A sequence converges if and only if it has exactly one limit point, and that limit point is the limit of the sequence.
• The set of limit points of any sequence is always a closed set.
• The largest limit point of a sequence is known as the limit superior (limsup), and the smallest limit point is known as the limit inferior (liminf). For this sequence, the limit superior is $\infty$ and the limit inferior is 0.
• The Bolzano-Weierstrass Theorem states that every bounded sequence in $\mathbb{R}$ has at least one limit point in $\mathbb{R}$. Our sequence is not bounded above, so this theorem doesn't directly apply to guarantee a limit point in $\mathbb{R}$, but our analysis showed 0 is a limit point.