Question

Determine the convergence or divergence of the sequence with the given \(n\) th term. If the sequence converges, find its limit. \(a_{n}=\frac{\ln \sqrt{n}}{n}\)

Short answer

The given sequence \(a_{n}=\frac{\ln{\sqrt{n}}}{n}\) converges and its limit is \(0\).

Step-by-step solution

Preliminary Mathematics

To check the convergence of \(a_{n}\), we need to find \(\lim_{{n \to \infty}}\frac{\ln \sqrt{n}}{n}\). Before we do this, a property of logarithms can be used that states \(\ln a^k = k \ln a\). Therefore, we can write the sequence as \(a_{n}=2\cdot\frac{\ln{n}}{n}\).

L'Hopital’s Rule

Next, it's important to note that this form is similar to \(\lim_{{x \to a}}\frac{f(x)}{g(x)}\), which is an indeterminate form of type 0/0 or ∞/∞. We can make use of L'Hopital’s rule here, which states that this limit is the same as the limit of the ratios of their derivatives. We differentiate the numerator and denominator separately, which gives us \(\frac{d}{d n}(\ln n) = 1/n\), \(\frac{d}{d n}n = 1\). Applying L'Hopital’s rule, we get \(\lim_{{n \to \infty}}\frac{\ln{n}}{n} = \lim_{{n \to \infty}}\frac{1/n}{1}\), which simplifies to \(\lim_{{n \to \infty}}(\frac{1}{n})\).

Finding the limit

Find the limit of \(\frac{1}{n}\) as \(n\) approaches infinity. The result is \(0\). Hence, the sequence \(a_{n}=\frac{\ln{\sqrt{n}}}{n}\) converges and its limit is \(0\).