Q. 48 - Calculus

Question

Determine whether the sequence is monotonic or eventually monotonic and whether the sequence is bounded above and/or below. If the sequence converges, give the limit. 

     $\left\{\frac{1\cdot 3\cdot 5\cdots (2k-1)}{1\cdot 4\cdot 7\cdots (3k-2)}\right\}$

Step-by-Step Solution

Answer

Ans:  The sequence$\left\{\frac{1\cdot 3\cdot 5\cdots (2k-1)}{1\cdot 4\cdot 7\cdots (3k-2)}\right\}$ is convergent and its limit is $0$.

Step 1. Given information.

given,

     $\left\{\frac{1\cdot 3\cdot 5\cdots (2k-1)}{1\cdot 4\cdot 7\cdots (3k-2)}\right\}$

Step 2. The objective is to determine whether the sequence is monotonic, bounded above, or bounded below, and to find the limit of the sequence if the sequence is convergent.

  The sequence $\left\{a_k\right\}=\left\{\frac{1\cdot 3\cdot 5\cdots (2k-1)}{1\cdot 4\cdot 7\cdots (3k-2)}\right\}$ the general term is  $a_k=\frac{1\cdot 3\cdot 5\cdots (2k-1)}{1\cdot 4\cdot 7\cdots (3k-2)}$.

Step 3. The general term of the sequence is a k = 1 ⋅ 3 ⋅ 5 ⋯ ( 2 k − 1 ) 1 ⋅ 4 ⋅ 7 ⋯ ( 3 k − 2 )

The ratio $\frac{a_{k+1}}{a_k}$ gives

 $\begin{array}{r}\frac{a_{k+1}}{a_k}=\frac{\frac{1\cdot 3\cdot 5\cdots (2k-1)(2k+1)}{1\cdot 4\cdot 7\cdots (3k-2)(3k+1)}}{\frac{1\cdot 3\cdot 5\cdots (2k-1)}{1\cdot 4\cdot 7\cdots (3k-2)}} \text{ (Substitution) }\\ =\frac{1\cdot 3\cdot 5\cdots (2k-1)(2k+1)}{1\cdot 4\cdot 7\cdots (3k-2)(3k+1)}\cdot \frac{1\cdot 4\cdot 7\cdots (3k-2)}{1\cdot 3\cdot 5\cdots (2k-1)} \\ =\frac{2k+1}{3k+1} \left(\text{ Simplify }\right)\\ <1\text{ (For }k>0) \end{array}$

Thus, $a_{k+1}<a_k$ when the value of $k>0$.

The sequence $\left\{a_k\right\}=\left\{\frac{1\cdot 3\cdot 5\cdots (2k-1)}{1\cdot 4\cdot 7\cdots (3k-2)}\right\}$ is eventually strictly increasing. The given sequence is monotonic.

Step 4. Now,

 The sequence $\left\{a_k\right\}=\left\{\frac{1\cdot 3\cdot 5\cdots (2k-1)}{1\cdot 4\cdot 7\cdots (3k-2)}\right\}$ is bounded below because

  $0<a_k$ for $k>0$

 As the index k increases, the term $\left\{a_k\right\}=\left\{\frac{1\cdot 3\cdot 5\cdots (2k-1)}{1\cdot 4\cdot 7\cdots (3k-2)}\right\}$ approaches $0$.

Thus, the strictly decreasing sequence has an upper bound $1$.

The given sequence has lower and upper bounds, therefore, the sequence is bounded. 

Step 5. The monotonic increasing sequence is bounded above is convergent

The strictly decreasing sequence $\left\{a_k\right\}=\left\{\frac{1\cdot 3\cdot 5\cdots (2k-1)}{1\cdot 4\cdot 7\cdots (3k-2)}\right\}$ is bounded below and hence is convergent. Therefore, the sequence is convergent.

Step 6. The limit of the sequence is { a k } = 1 &#8901; 3 &#8901; 5 &#8943; ( 2 k &#8722; 1 ) 1 &#8901; 4 &#8901; 7 &#8943; ( 3 k &#8722; 2 )

  $\begin{array}{r}\underset{k\to \mathrm{\infty }}{lim} a_k=\underset{k\to \mathrm{\infty }}{lim} \left(\frac{1\cdot 3\cdot 5\cdots (2k-1)}{1\cdot 4\cdot 7\cdots (3k-2)}\right)\\ =0 \left(\text{ Simplify }\right)\end{array}$

Thus, the limit of the sequence $\left\{a_k\right\}=\left\{\frac{1\cdot 3\cdot 5\cdots (2k-1)}{1\cdot 4\cdot 7\cdots (3k-2)}\right\}$ is $0$.