The problem asks us to analyze several sequences given by their explicit formulas. For each sequence $a_n$, we need to find the first five terms, determine if the sequence converges or diverges, and if it converges, find the limit as $n$ approaches infinity. I will analyze problems 1, 3, 5, 6, 11, 13, 14, 16, 17 and 18.

1. Problem Description

The problem asks us to analyze several sequences given by their explicit formulas. For each sequence $a_n$, we need to find the first five terms, determine if the sequence converges or diverges, and if it converges, find the limit as $n$ approaches infinity. I will analyze problems 1, 3, 5, 6, 11, 13, 14, 16, 17 and

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2. Solution Steps

Problem 1: $a_n = \frac{n}{3n-1}$

First five terms:

$a_1 = \frac{1}{3(1)-1} = \frac{1}{2}$

$a_2 = \frac{2}{3(2)-1} = \frac{2}{5}$

$a_3 = \frac{3}{3(3)-1} = \frac{3}{8}$

$a_4 = \frac{4}{3(4)-1} = \frac{4}{11}$

$a_5 = \frac{5}{3(5)-1} = \frac{5}{14}$

To find the limit, divide both numerator and denominator by $n$:

$\lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac{n}{3n-1} = \lim_{n \to \infty} \frac{1}{3 - \frac{1}{n}} = \frac{1}{3-0} = \frac{1}{3}$

The sequence converges to $\frac{1}{3}$.

Problem 3: $a_n = \frac{4n^2+2}{n^2+3n-1}$

First five terms:

$a_1 = \frac{4(1)^2+2}{1^2+3(1)-1} = \frac{6}{3} = 2$

$a_2 = \frac{4(2)^2+2}{2^2+3(2)-1} = \frac{18}{9} = 2$

$a_3 = \frac{4(3)^2+2}{3^2+3(3)-1} = \frac{38}{17}$

$a_4 = \frac{4(4)^2+2}{4^2+3(4)-1} = \frac{66}{27} = \frac{22}{9}$

$a_5 = \frac{4(5)^2+2}{5^2+3(5)-1} = \frac{102}{39} = \frac{34}{13}$

To find the limit, divide both numerator and denominator by $n^2$:

$\lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac{4n^2+2}{n^2+3n-1} = \lim_{n \to \infty} \frac{4 + \frac{2}{n^2}}{1 + \frac{3}{n} - \frac{1}{n^2}} = \frac{4+0}{1+0-0} = 4$

The sequence converges to $4$.

Problem 5: $a_n = \frac{n^3 + 3n^2 + 3n}{(n+1)^3}$

First five terms:

$a_1 = \frac{1^3 + 3(1)^2 + 3(1)}{(1+1)^3} = \frac{7}{8}$

$a_2 = \frac{2^3 + 3(2)^2 + 3(2)}{(2+1)^3} = \frac{8+12+6}{27} = \frac{26}{27}$

$a_3 = \frac{3^3 + 3(3)^2 + 3(3)}{(3+1)^3} = \frac{27+27+9}{64} = \frac{63}{64}$

$a_4 = \frac{4^3 + 3(4)^2 + 3(4)}{(4+1)^3} = \frac{64+48+12}{125} = \frac{124}{125}$

$a_5 = \frac{5^3 + 3(5)^2 + 3(5)}{(5+1)^3} = \frac{125+75+15}{216} = \frac{215}{216}$

$a_n = \frac{n^3 + 3n^2 + 3n}{(n+1)^3} = \frac{n^3 + 3n^2 + 3n}{n^3 + 3n^2 + 3n + 1}$

$\lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac{n^3 + 3n^2 + 3n}{n^3 + 3n^2 + 3n + 1} = \lim_{n \to \infty} \frac{1 + \frac{3}{n} + \frac{3}{n^2}}{1 + \frac{3}{n} + \frac{3}{n^2} + \frac{1}{n^3}} = \frac{1+0+0}{1+0+0+0} = 1$

The sequence converges to $1$.

Problem 6: $a_n = \frac{\sqrt{3n^2 + 2}}{2n + 1}$

First five terms:

$a_1 = \frac{\sqrt{3(1)^2+2}}{2(1)+1} = \frac{\sqrt{5}}{3}$

$a_2 = \frac{\sqrt{3(2)^2+2}}{2(2)+1} = \frac{\sqrt{14}}{5}$

$a_3 = \frac{\sqrt{3(3)^2+2}}{2(3)+1} = \frac{\sqrt{29}}{7}$

$a_4 = \frac{\sqrt{3(4)^2+2}}{2(4)+1} = \frac{\sqrt{50}}{9} = \frac{5\sqrt{2}}{9}$

$a_5 = \frac{\sqrt{3(5)^2+2}}{2(5)+1} = \frac{\sqrt{77}}{11}$

$\lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac{\sqrt{3n^2 + 2}}{2n + 1} = \lim_{n \to \infty} \frac{\sqrt{n^2(3 + \frac{2}{n^2})}}{n(2 + \frac{1}{n})} = \lim_{n \to \infty} \frac{n\sqrt{3 + \frac{2}{n^2}}}{n(2 + \frac{1}{n})} = \lim_{n \to \infty} \frac{\sqrt{3 + \frac{2}{n^2}}}{2 + \frac{1}{n}} = \frac{\sqrt{3+0}}{2+0} = \frac{\sqrt{3}}{2}$

The sequence converges to $\frac{\sqrt{3}}{2}$.

Problem 11: $a_n = \frac{e^{2n}}{n^2 + 3n - 1}$

First five terms:

$a_1 = \frac{e^{2(1)}}{1^2 + 3(1) - 1} = \frac{e^2}{3}$

$a_2 = \frac{e^{2(2)}}{2^2 + 3(2) - 1} = \frac{e^4}{9}$

$a_3 = \frac{e^{2(3)}}{3^2 + 3(3) - 1} = \frac{e^6}{17}$

$a_4 = \frac{e^{2(4)}}{4^2 + 3(4) - 1} = \frac{e^8}{27}$

$a_5 = \frac{e^{2(5)}}{5^2 + 3(5) - 1} = \frac{e^{10}}{39}$

Using L'Hopital's rule:

$\lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac{e^{2n}}{n^2 + 3n - 1} = \lim_{n \to \infty} \frac{2e^{2n}}{2n + 3} = \lim_{n \to \infty} \frac{4e^{2n}}{2} = \lim_{n \to \infty} 2e^{2n} = \infty$

The sequence diverges to $\infty$.

Problem 13: $a_n = \frac{(-\pi)^n}{5^n}$

First five terms:

$a_1 = \frac{(-\pi)^1}{5^1} = -\frac{\pi}{5}$

$a_2 = \frac{(-\pi)^2}{5^2} = \frac{\pi^2}{25}$

$a_3 = \frac{(-\pi)^3}{5^3} = -\frac{\pi^3}{125}$

$a_4 = \frac{(-\pi)^4}{5^4} = \frac{\pi^4}{625}$

$a_5 = \frac{(-\pi)^5}{5^5} = -\frac{\pi^5}{3125}$

$a_n = \left(-\frac{\pi}{5}\right)^n$

Since $|\frac{-\pi}{5}| = \frac{\pi}{5} \approx \frac{3.14}{5} < 1$, the sequence converges to $0$.

$\lim_{n \to \infty} a_n = \lim_{n \to \infty} \left(-\frac{\pi}{5}\right)^n = 0$

Problem 14: $a_n = (\frac{1}{4})^n + \frac{3n}{2}$

First five terms:

$a_1 = (\frac{1}{4})^1 + \frac{3(1)}{2} = \frac{1}{4} + \frac{3}{2} = \frac{1}{4} + \frac{6}{4} = \frac{7}{4}$

$a_2 = (\frac{1}{4})^2 + \frac{3(2)}{2} = \frac{1}{16} + 3 = \frac{1+48}{16} = \frac{49}{16}$

$a_3 = (\frac{1}{4})^3 + \frac{3(3)}{2} = \frac{1}{64} + \frac{9}{2} = \frac{1+288}{64} = \frac{289}{64}$

$a_4 = (\frac{1}{4})^4 + \frac{3(4)}{2} = \frac{1}{256} + 6 = \frac{1+1536}{256} = \frac{1537}{256}$

$a_5 = (\frac{1}{4})^5 + \frac{3(5)}{2} = \frac{1}{1024} + \frac{15}{2} = \frac{1+7680}{1024} = \frac{7681}{1024}$

$\lim_{n \to \infty} a_n = \lim_{n \to \infty} (\frac{1}{4})^n + \frac{3n}{2} = 0 + \infty = \infty$

The sequence diverges to $\infty$.

Problem 16: $a_n = \frac{n^{100}}{e^n}$

First five terms:

$a_1 = \frac{1^{100}}{e^1} = \frac{1}{e}$

$a_2 = \frac{2^{100}}{e^2}$

$a_3 = \frac{3^{100}}{e^3}$

$a_4 = \frac{4^{100}}{e^4}$

$a_5 = \frac{5^{100}}{e^5}$

Using L'Hopital's rule repeatedly, we know that exponential growth is faster than polynomial growth.

$\lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac{n^{100}}{e^n} = 0$

The sequence converges to $0$.

Problem 17: $a_n = \frac{\ln n}{\sqrt{n}}$

First five terms:

$a_1 = \frac{\ln 1}{\sqrt{1}} = \frac{0}{1} = 0$

$a_2 = \frac{\ln 2}{\sqrt{2}} \approx \frac{0.693}{1.414} \approx 0.49$

$a_3 = \frac{\ln 3}{\sqrt{3}} \approx \frac{1.099}{1.732} \approx 0.63$

$a_4 = \frac{\ln 4}{\sqrt{4}} = \frac{2\ln 2}{2} = \ln 2 \approx 0.693$

$a_5 = \frac{\ln 5}{\sqrt{5}} \approx \frac{1.609}{2.236} \approx 0.72$

Using L'Hopital's rule:

$\lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac{\ln n}{\sqrt{n}} = \lim_{n \to \infty} \frac{\frac{1}{n}}{\frac{1}{2\sqrt{n}}} = \lim_{n \to \infty} \frac{2\sqrt{n}}{n} = \lim_{n \to \infty} \frac{2}{\sqrt{n}} = 0$

The sequence converges to $0$.

Problem 18: $a_n = \frac{\ln(\frac{1}{n})}{\sqrt{2n}}$

First five terms:

$a_1 = \frac{\ln(\frac{1}{1})}{\sqrt{2(1)}} = \frac{\ln 1}{\sqrt{2}} = \frac{0}{\sqrt{2}} = 0$

$a_2 = \frac{\ln(\frac{1}{2})}{\sqrt{2(2)}} = \frac{-\ln 2}{2} \approx \frac{-0.693}{2} \approx -0.347$

$a_3 = \frac{\ln(\frac{1}{3})}{\sqrt{2(3)}} = \frac{-\ln 3}{\sqrt{6}} \approx \frac{-1.099}{2.449} \approx -0.449$

$a_4 = \frac{\ln(\frac{1}{4})}{\sqrt{2(4)}} = \frac{-\ln 4}{\sqrt{8}} = \frac{-2\ln 2}{2\sqrt{2}} = \frac{-\ln 2}{\sqrt{2}} \approx -0.49$

$a_5 = \frac{\ln(\frac{1}{5})}{\sqrt{2(5)}} = \frac{-\ln 5}{\sqrt{10}} \approx \frac{-1.609}{3.162} \approx -0.509$

$\lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac{\ln(\frac{1}{n})}{\sqrt{2n}} = \lim_{n \to \infty} \frac{-\ln n}{\sqrt{2n}}$

Using L'Hopital's rule:

$\lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac{-\ln n}{\sqrt{2n}} = \lim_{n \to \infty} \frac{-\frac{1}{n}}{\frac{\sqrt{2}}{2\sqrt{n}}} = \lim_{n \to \infty} \frac{-2\sqrt{n}}{n\sqrt{2}} = \lim_{n \to \infty} \frac{-2}{\sqrt{2n}} = 0$

The sequence converges to $0$.

3. Final Answer

Problem 1: First five terms are $\frac{1}{2}, \frac{2}{5}, \frac{3}{8}, \frac{4}{11}, \frac{5}{14}$. The sequence converges to $\frac{1}{3}$.

Problem 3: First five terms are $2, 2, \frac{38}{17}, \frac{22}{9}, \frac{34}{13}$. The sequence converges to $4$.

Problem 5: First five terms are $\frac{7}{8}, \frac{26}{27}, \frac{63}{64}, \frac{124}{125}, \frac{215}{216}$. The sequence converges to $1$.

Problem 6: First five terms are $\frac{\sqrt{5}}{3}, \frac{\sqrt{14}}{5}, \frac{\sqrt{29}}{7}, \frac{5\sqrt{2}}{9}, \frac{\sqrt{77}}{11}$. The sequence converges to $\frac{\sqrt{3}}{2}$.

Problem 11: First five terms are $\frac{e^2}{3}, \frac{e^4}{9}, \frac{e^6}{17}, \frac{e^8}{27}, \frac{e^{10}}{39}$. The sequence diverges to $\infty$.

Problem 13: First five terms are $-\frac{\pi}{5}, \frac{\pi^2}{25}, -\frac{\pi^3}{125}, \frac{\pi^4}{625}, -\frac{\pi^5}{3125}$. The sequence converges to $0$.

Problem 14: First five terms are $\frac{7}{4}, \frac{49}{16}, \frac{289}{64}, \frac{1537}{256}, \frac{7681}{1024}$. The sequence diverges to $\infty$.

Problem 16: First five terms are $\frac{1}{e}, \frac{2^{100}}{e^2}, \frac{3^{100}}{e^3}, \frac{4^{100}}{e^4}, \frac{5^{100}}{e^5}$. The sequence converges to $0$.

Problem 17: First five terms are $0, \frac{\ln 2}{\sqrt{2}}, \frac{\ln 3}{\sqrt{3}}, \ln 2, \frac{\ln 5}{\sqrt{5}}$. The sequence converges to $0$.

Problem 18: First five terms are $0, \frac{-\ln 2}{2}, \frac{-\ln 3}{\sqrt{6}}, \frac{-\ln 2}{\sqrt{2}}, \frac{-\ln 5}{\sqrt{10}}$. The sequence converges to $0$.