Sequences — Question 7

Question 7 – Solution

Step 1: Choose a continuous extension.

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 $a_n\to0$.

Step 2: Compare possible methods.

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.

Step 3: Determine where the sequence increases and decreases.

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.