Sequences — Question 6

Question 6 – Solution

Step 1: Transform the variable exponent.

Because $a_n=n^{1/n}>0$, logarithms are valid. Let $b_n=\ln a_n$. Then $$b_n=\ln(n^{1/n})=\frac{\ln n}{n}.$$

Step 2: Evaluate the logarithmic limit.

Extend the quotient to $g(x)=\ln x/x$. It has the indeterminate form $\infty/\infty$, and L’Hopital’s Rule gives $$\lim_{x\to\infty}\frac{\ln x}{x}=\lim_{x\to\infty}\frac{1/x}{1}=0.$$ Restricting this continuous limit to integer values gives $b_n\to0$. By continuity of the exponential function, $a_n=e^{b_n}\to e^0=1$.

Step 3: Determine when the terms exceed the limit.

For every integer $n>1$, the base $n>1$ and exponent $1/n>0$, so $a_n>1$; also $a_1=1$. A sequence may approach a limit entirely from one side, so convergence to $1$ does not require later terms to equal or cross $1$.

Step 4: Explain the rise and fall.

For the continuous extension $f(x)=x^{1/x}$, logarithmic differentiation gives $$\frac{f'(x)}{f(x)}=\frac{1-\ln x}{x^2}.$$ Because $f(x)>0$, the sign of $f'(x)$ is the sign of $1-\ln x$. Thus $f$ increases for $x<e$ and decreases for $x>e$. Among integer indices, the maximum occurs at $n=3$, after which the sequence decreases toward $1$ while remaining above it.