Arithmetic Sequence and Series Limit Calculation

This problem requires calculating the limit of an infinite series based on an arithmetic sequence. We will identify the sequence, analyze the general term of the series, perform a partial fraction decomposition, and then evaluate the telescoping sum and its limit.

Identifying the Arithmetic Sequence

The given sequence is defined by $a_1 = 1$ and $a_n = a_{n-1} + 4$ for $n \ge 2$. This describes an arithmetic progression with:

• First term: $a_1 = 1$
• Common difference: $d = 4$

The formula for the $n^{th}$ term of an arithmetic progression is $a_n = a_1 + (n-1)d$. Substituting the values:

$a_n = 1 + (n-1)4 = 1 + 4n - 4 = 4n - 3$

Analyzing the Series Term

The series is $\frac{1}{a_1a_2} + \frac{1}{a_2a_3} + \dots + \frac{1}{a_{n-1}a_n}$. The general term within the summation is $\frac{1}{a_{k-1}a_k}$. Using the formula $a_n = 4n - 3$, we find $a_{k-1} = 4(k-1) - 3 = 4k - 7$ and $a_k = 4k - 3$.

We apply partial fraction decomposition to the term $\frac{1}{a_{k-1}a_k} = \frac{1}{(4k-7)(4k-3)}$. The difference between the factors in the denominator is $(4k-3) - (4k-7) = 4$. Therefore:

$ \frac{1}{a_{k-1}a_k} = \frac{1}{(4k-7)(4k-3)} = \frac{1}{4} \left( \frac{1}{4k-7} - \frac{1}{4k-3} \right) $

Replacing the factors with the sequence terms:

$ \frac{1}{a_{k-1}a_k} = \frac{1}{4} \left( \frac{1}{a_{k-1}} - \frac{1}{a_k} \right) $

Calculating the Partial Sum

Let $S_n$ represent the sum of the first $n-1$ product terms in the denominator, which is the sum up to $\frac{1}{a_{n-1}a_n}$. The sum is:

$ S_n = \sum_{k=2}^{n} \frac{1}{a_{k-1}a_k} = \sum_{k=2}^{n} \frac{1}{4} \left( \frac{1}{a_{k-1}} - \frac{1}{a_k} \right) $

This summation is a telescoping series. Expanding the terms reveals cancellation:

$ S_n = \frac{1}{4} \left[ \left( \frac{1}{a_1} - \frac{1}{a_2} \right) + \left( \frac{1}{a_2} - \frac{1}{a_3} \right) + \dots + \left( \frac{1}{a_{n-1}} - \frac{1}{a_n} \right) \right] $

After cancellation, the sum simplifies to:

$ S_n = \frac{1}{4} \left( \frac{1}{a_1} - \frac{1}{a_n} \right) $

Computing the Limit

To find the limit of the series, we evaluate $S_n$ as $n$ approaches infinity:

$ \lim_{n\to\infty} S_n = \lim_{n\to\infty} \frac{1}{4} \left( \frac{1}{a_1} - \frac{1}{a_n} \right) $

Given $a_1 = 1$ and $a_n = 4n - 3$. As $n \to \infty$, $a_n$ grows infinitely large ($a_n \to \infty$), so $\frac{1}{a_n} \to 0$.

Substituting these values:

$ \lim_{n\to\infty} S_n = \frac{1}{4} \left( \frac{1}{1} - 0 \right) = \frac{1}{4} \times 1 = \frac{1}{4} $

The calculated value is $\frac{1}{4} = 0.25$. This value is between $0.24$ and $0.26$.