Sequences — Question 7

*Question 7 Let $a_n=\dfrac{\ln n}{\sqrt n}$ for $n\ge1$.

1. Use a continuous extension and L’Hopital’s Rule to find the limit.

2. Explain why taking logarithms of $a_n$ is not the most direct method.

3. Use the derivative of $f(x)=\ln x/\sqrt{x}$ to explain the initial increase of the sequence and identify where the eventual decrease begins.

*Question 7 – Solution

Let $f(x)=\ln x/\sqrt{x}$ for $x\ge1$. Both numerator and denominator tend to infinity, so the quotient has the indeterminate form $\infty/\infty$ and L’Hopital’s Rule applies: $$\lim_{x\to\infty}\frac{\ln x}{\sqrt{x}} =\lim_{x\to\infty}\frac{1/x}{1/(2\sqrt{x})} =\lim_{x\to\infty}\frac2{\sqrt{x}}=0.$$ Restricting the continuous limit to integer inputs gives $a_n\to0$.

Taking $\ln(a_n)=\ln(\ln n)-\tfrac12\ln n$ would introduce an $\infty-\infty$ expression and still require comparing logarithmic growth. Applying L’Hopital’s Rule directly to the original quotient is cleaner.

Differentiate the extension: $$f'(x)=x^{-3/2}\left(1-\frac12\ln x\right).$$ Since $x^{-3/2}>0$, $f'(x)>0$ for $x<e^2$, equals $0$ at $e^2\approx7.39$, and is negative afterward. The integer terms rise through $n=7$ and then decrease very slowly. This initial increase is consistent with the eventual limit of zero.