find the n th term of the sequence 3,2,5/3,6/4, 7/5,.. and find its limit if exist

To find the \( n \)-th term of the sequence, we will first observe the general pattern.

Given sequence: 
\[
3, 2, \frac{5}{3}, \frac{6}{4}, \frac{7}{5}, \ldots
\]

The numerators of the terms are increasing by 1, starting from 3: 
\[ 3, 2, 5, 6, 7, \ldots \]
which suggests that the numerator follows the pattern \( 3 + (n - 1) = 2 + n \).

The denominators follow the sequence: 
\[ 1, 1, 3, 4, 5, \ldots \]
which suggests that the denominator for the \( n \)-th term is \( n + 1 \).

Thus, the general form of the \( n \)-th term \( a_n \) is:
\[
a_n = \frac{2 + n}{n + 1}
\]

### Finding the limit of the sequence:

Now, we compute the limit of the sequence as \( n \to \infty \):

\[
\lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac{2 + n}{n + 1}
\]

We can divide both the numerator and denominator by \( n \):

\[
\lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac{\frac{2}{n} + 1}{1 + \frac{1}{n}}
\]

As \( n \to \infty \), both \( \frac{2}{n} \to 0 \) and \( \frac{1}{n} \to 0 \). Therefore, we get:

\[
\lim_{n \to \infty} a_n = \frac{0 + 1}{1 + 0} = 1
\]

So, the limit of the sequence is 1 as \( n \to \infty \).

### Summary:

- The \( n \)-th term of the sequence is \( \frac{2 + n}{n + 1} \).
- The limit of the sequence as \( n \to \infty \) is 1.

Do you need further clarification on this? 

Here are 5 related questions:
1. Can you determine the general form of a sequence from its first few terms?
2. How do you find the limit of more complex sequences?
3. What is the difference between arithmetic and geometric sequences?
4. How can you determine whether a sequence converges or diverges?
5. How does L'Hopital's rule help in finding the limit of sequences?

**Tip:** When dealing with sequences, identifying the pattern in the numerators and denominators separately can often help to find the general term.

Find the n-th term of the sequence 3, 2, 5/3, 6/4, 7/5,... and find its limit if it exists.

This detailed guide explains how to find the n-th term and limit of the sequence 3, 2, 5/3, 6/4, 7/5,... The solution covers the algebraic pattern of the sequence, the formula for the general term, and how to calculate the limit as n approaches infinity. Suitable for students in Grades 9-12.

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