91Ó°ÊÓ

To prove the inequality \(a_1 < a_2 < a_3\), we will compute the values of each term in the sequence. For \(n = 1\), we have \(a_1 = 1^{\frac{1}{1}} = 1\). For \(n = 2\), we have \(a_2 = 2^{\frac{1}{2}} = \sqrt{2}\). For \(n = 3\), we have \(a_3 = 3^{\frac{1}{3}} \approx 1.4422\). Thus, we can clearly see that \(1 < \sqrt{2} < 1.4422\), which establishes that \(a_1 < a_2 < a_3\).

To show that \(a_n > a_{n+1}\) for all \(n\geq3\), we will prove that \(n^{\frac{1}{n}} > (n+1)^{\frac{1}{n+1}}\) by taking the logarithm of both sides. Recall that \(\ln x\) is a monotonically increasing function. Thus, if \(\ln(a_n) > \ln(a_{n+1})\), we can deduce that \(a_n > a_{n+1}\). Consider the function \(f(x) = \ln x - \frac{x-1}{x}\ln(n+1)\). Now, we need to find \(f'(x)\). \(f'(x) = \frac{1}{x} - \frac{\ln(n+1)}{x^2}\) Let's find the critical points of \(f'(x)\) by setting \(f'(x)=0\): \(\frac{1}{x} - \frac{\ln(n+1)}{x^2} = 0\) Solving for x, we have \(x = \frac{\ln(n+1)}{x}\). Now, using the second derivative test, we find \(f''(x)\): \(f''(x) = -\frac{1}{x^2} + 2\frac{\ln(n+1)}{x^3}\) Since \(x > 0\) and \(\ln(n+1) > 0\), \(f''(x) > 0\). Thus, the function \(f(x)\) is concave up and has a unique global minimum at \(x = \frac{\ln(n+1)}{x}\). Now, if we choose \(x = n\) and \(x = n+1\) such that \(x \in [n, n+1]\): $$f(n) = \ln(n) - \frac{n-1}{n}\ln(n+1)$$ $$f(n+1) = \ln(n+1) - \frac{n}{n+1}\ln(n+1)$$ Since \(f(x)\) is concave up and \(n < \frac{\ln(n+1)}{x} < n+1\), we have \(f(n) > f(n+1)\). Thus, \(\ln a_n > \ln a_{n+1}\) and hence, \(a_n > a_{n+1}\) for all \(n \geq 3\).

It is given that, $$1 < a_n < 1 + \frac{\sqrt{2}}{\sqrt{n-1}}$$ for all \(n \geq 2\). Recall that \(a_n = n^{\frac{1}{n}}\). Now, observe that for \(n = 2\), the inequality is true. We could also use the fact that \(a_n > a_{n+1}\) and consider their limits to deduce that the sequence converges to 1 as \(n \to \infty\).

To show that the limit of the sequence \(a_n\) is 1 as \(n\) approaches infinity, we can use the squeeze theorem. Since \(a_{n} > a_{n+1}\) for all \(n \geq 3\) and \(1 < a_n < 1 + \frac{\sqrt{2}}{\sqrt{n-1}}\) for all \(n \geq 2\), we can say that \(a_n\) is a decreasing sequence. Also, note that \(a_n > 0\) for all \(n \geq 1\) and it is bounded below by 1. Thus, the sequence \(a_n\) is a decreasing sequence, bounded below by 1. So, there exists a limit as n approaches infinity. By using the squeeze theorem (sandwich theorem), the limit of \(a_n\) as \(n\) approaches infinity will be equal to 1.