The problem asks us to find an explicit formula $a_n$ for the given sequence, determine whether the sequence converges or diverges, and if it converges, find the limit of $a_n$ as $n$ approaches infinity. The sequence is given as $1, \frac{2}{2^2 - 1^2}, \frac{3}{3^2 - 2^2}, \frac{4}{4^2 - 3^2}, \dots$

1. Problem Description

The problem asks us to find an explicit formula $a_n$ for the given sequence, determine whether the sequence converges or diverges, and if it converges, find the limit of $a_n$ as $n$ approaches infinity. The sequence is given as $1, \frac{2}{2^2 - 1^2}, \frac{3}{3^2 - 2^2}, \frac{4}{4^2 - 3^2}, \dots$

2. Solution Steps

First, let's find the explicit formula $a_n$ for the sequence. The first term is

1. For $n \ge 2$, the general term can be written as:

$a_n = \frac{n}{n^2 - (n-1)^2}$.

Simplifying the denominator:

$n^2 - (n-1)^2 = n^2 - (n^2 - 2n + 1) = n^2 - n^2 + 2n - 1 = 2n - 1$.

Thus, for $n \ge 2$, $a_n = \frac{n}{2n - 1}$.

We can write the explicit formula as:

$a_1 = 1$ and $a_n = \frac{n}{2n - 1}$ for $n \ge 2$.

To see if a single formula can be used for all $n \ge 1$, let's check if $a_1$ matches.

For $n=1$, we would have $a_1 = \frac{1}{2(1) - 1} = \frac{1}{1} = 1$, which matches the first term of the sequence.

So, the explicit formula for the sequence is $a_n = \frac{n}{2n - 1}$ for $n \ge 1$.

Next, we determine whether the sequence converges or diverges. We need to find the limit of $a_n$ as $n$ approaches infinity:

$\lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac{n}{2n - 1}$.

We can divide both the numerator and the denominator by $n$:

$\lim_{n \to \infty} \frac{n/n}{(2n - 1)/n} = \lim_{n \to \infty} \frac{1}{2 - \frac{1}{n}}$.

As $n \to \infty$, $\frac{1}{n} \to 0$, so we have:

$\lim_{n \to \infty} \frac{1}{2 - \frac{1}{n}} = \frac{1}{2 - 0} = \frac{1}{2}$.

Since the limit exists and is equal to $\frac{1}{2}$, the sequence converges.

3. Final Answer

$a_n = \frac{n}{2n - 1}$

The sequence converges.

$\lim_{n \to \infty} a_n = \frac{1}{2}$