The problem requires us to find the first five terms of the given sequences $\{a_n\}$, determine if each sequence converges or diverges, and if it converges, find its limit as $n$ approaches infinity. We will solve problem number 2: $a_n = \frac{3n+2}{n+1}$.

AnalysisSequencesLimitsConvergenceDivergence
2025/5/27

1. Problem Description

The problem requires us to find the first five terms of the given sequences {an}\{a_n\}, determine if each sequence converges or diverges, and if it converges, find its limit as nn approaches infinity. We will solve problem number 2: an=3n+2n+1a_n = \frac{3n+2}{n+1}.

2. Solution Steps

First, we find the first five terms of the sequence by substituting n=1,2,3,4,5n=1, 2, 3, 4, 5 into the formula for ana_n:
a1=3(1)+21+1=52=2.5a_1 = \frac{3(1)+2}{1+1} = \frac{5}{2} = 2.5
a2=3(2)+22+1=832.67a_2 = \frac{3(2)+2}{2+1} = \frac{8}{3} \approx 2.67
a3=3(3)+23+1=114=2.75a_3 = \frac{3(3)+2}{3+1} = \frac{11}{4} = 2.75
a4=3(4)+24+1=145=2.8a_4 = \frac{3(4)+2}{4+1} = \frac{14}{5} = 2.8
a5=3(5)+25+1=1762.83a_5 = \frac{3(5)+2}{5+1} = \frac{17}{6} \approx 2.83
Next, we find the limit of the sequence as nn approaches infinity:
limnan=limn3n+2n+1\lim_{n\to\infty} a_n = \lim_{n\to\infty} \frac{3n+2}{n+1}
To find this limit, we can divide both the numerator and the denominator by nn:
limn3n+2n+1=limn3nn+2nnn+1n=limn3+2n1+1n\lim_{n\to\infty} \frac{3n+2}{n+1} = \lim_{n\to\infty} \frac{\frac{3n}{n}+\frac{2}{n}}{\frac{n}{n}+\frac{1}{n}} = \lim_{n\to\infty} \frac{3+\frac{2}{n}}{1+\frac{1}{n}}
As nn approaches infinity, 2n\frac{2}{n} and 1n\frac{1}{n} both approach

0. Therefore,

limn3+2n1+1n=3+01+0=31=3\lim_{n\to\infty} \frac{3+\frac{2}{n}}{1+\frac{1}{n}} = \frac{3+0}{1+0} = \frac{3}{1} = 3
Since the limit exists and is equal to 3, the sequence converges to
3.

3. Final Answer

The first five terms of the sequence are 2.5,83,114,145,1762.5, \frac{8}{3}, \frac{11}{4}, \frac{14}{5}, \frac{17}{6}.
The sequence converges, and limnan=3\lim_{n\to\infty} a_n = 3.

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