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. If the nth term of the sequence < an > is given by"211 322 433· ·n + 1nn# 1nFind the limit of sequence

Question

If the nth term of the sequence <an> < a_n > is given by:

an=21imes1+32imes2+43imes3++n+1×n a_n = 21 imes 1 + 32 imes 2 + 43 imes 3 + \cdots + n + 1 \times n

Find the limit of sequence.

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Solution

1. Break Down the Problem

We need to find the limit of the sequence defined by the nth term an a_n , which is given as: an=211322433(n+1)nn a_n = \frac{21}{1} \cdot \frac{32}{2} \cdot \frac{43}{3} \cdots \frac{(n+1)n}{n}

2. Relevant Concepts

To analyze this sequence, we can express each term in a more simplified format. The general term appears to be: (n+1)(n)(n1)(2)n! \frac{(n+1)(n)(n-1)\ldots(2)}{n!} We need to simplify an a_n and investigate its behavior as n n approaches infinity.

3. Analysis and Detail

Let’s write down the nth term clearly: an=k=1n(k+1)kk=k=1n(k+1)=(n+1)! a_n = \prod_{k=1}^{n} \frac{(k+1)k}{k} = \prod_{k=1}^{n} (k+1) = (n+1)! Thus, we have: an=(n+1)! a_n = (n+1)! Next, we need to find the limit of an a_n as n n approaches infinity.

4. Verify and Summarize

We know that as n n approaches infinity, (n+1)! (n+1)! grows without bound. Therefore, the limit can be mathematically stated as: limnan=limn(n+1)!= \lim_{n \to \infty} a_n = \lim_{n \to \infty} (n+1)! = \infty Thus, the sequence diverges to infinity.

Final Answer

The limit of the sequence an a_n as n n approaches infinity is: \infty

This problem has been solved

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