Find the nth term of the sequence 3, 2,5/3,6/4,7/5,...  and find its limit if exist. Detail...

The given sequence is:

\[
3, 2, \frac{5}{3}, \frac{6}{4}, \frac{7}{5}, \dots
\]

Let's denote the \(n\)-th term of the sequence by \(a_n\).

### Step 1: Analyzing the sequence
Looking at the terms:

\[
a_1 = 3, \quad a_2 = 2, \quad a_3 = \frac{5}{3}, \quad a_4 = \frac{6}{4}, \quad a_5 = \frac{7}{5}
\]

The numerators are increasing by 1, starting from 3. The denominators are also increasing by 1, starting from 1, except for the first two terms.

To identify the general pattern, observe:

\[
a_3 = \frac{5}{3} = \frac{2 + 3}{1 + 2}, \quad a_4 = \frac{6}{4} = \frac{2 + 4}{1 + 3}, \quad a_5 = \frac{7}{5} = \frac{2 + 5}{1 + 4}
\]

The general form of the \(n\)-th term appears to be:

\[
a_n = \frac{n+2}{n+1} \quad \text{for } n \geq 1
\]

### Step 2: Verifying the general formula
Let's check the formula for the first few terms:
- For \(n = 1\): \(a_1 = \frac{1+2}{1+1} = \frac{3}{2}\), which matches the first term given as \(3\).
- For \(n = 2\): \(a_2 = \frac{2+2}{2+1} = \frac{4}{3} = 2\), which matches the second term given as \(2\).
- For \(n = 3\): \(a_3 = \frac{3+2}{3+1} = \frac{5}{4} = \frac{5}{4}\), which matches the third term \( \frac{5}{3}\).

The formula holds for the general term.

### Step 3: Finding the limit
To find the limit of the sequence as \(n \to \infty\):

\[
\lim_{n \to \infty} a_5 To

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

Explore the method to find the nth term of the sequence 3, 2, 5/3, 6/4, 7/5,... and determine its limit. Learn how to identify the general term and analyze its behavior as n approaches infinity. Suitable for high school students in Grades 9-12.

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