$\lim_{n\to\infty} \left[\frac{1}{a_1a_2} + \frac{1}{a_2a_3} + \dots + \frac{1}{a_{n-1}a_n}\right]$
is equal to ________
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 $0.26$.
Important Questions from Series
- In the sequence 6, 9, 14, $x$, 30, 41, a possible value of $x$ is
Let $a_0 = 0$ and define $a_n = \frac{1}{2}(1 + a_{n-1})$ for all positive integers $n \ge 1$.
The least value of $n$ for which $|1 - a_n| < \frac{1}{2^{10}}$ is __________.
(Answer in integer)
- Calculate the reciprocal of the coefficient of $z^3$ in the Taylor series expansion of the function $f(z) = \sin(z)$ around $z = 0$. (Provide the answer as an integer.)
- Let $S(x) = a_0 + \sum_{n=1}^\infty(a_n \cos (n x) + b_n \sin (n x))$ be the Fourier series of the$2 \pi$ periodic function defined by $f(x) = x^2 + 4 \sin (x) \cos(x)$, $-\pi \le x \le \pi$. Then
$|\sum_{n=0}^\infty a_n - \sum_{n=1}^\infty b_n|$
is equal to ________ Let $S_n = \sum_{k=1}^n \frac{1}{k}$ and $I_n = \int_1^n \frac{x - [x]}{x^2} dx$. Then, $S_{10} + I_{10}$ is equal to
