009C Sample Midterm 3, Problem 1 Detailed Solution

Test if the following sequence ${a_{n}}$ converges or diverges.

If it converges, also find the limit of the sequence.

$a_{n}=\left({\frac {n-7}{n}}\right)^{\frac {1}{n}}$ Background Information:
L'Hôpital's Rule, Part 2

Let  $f$ and  $g$ be differentiable functions on the open interval  $(a,\infty )$ for some value  $a,$ where  $g'(x)\neq 0$ on  $(a,\infty )$ and  $\lim _{x\rightarrow \infty }{\frac {f(x)}{g(x)}}$ returns either  ${\frac {0}{0}}$ or  ${\frac {\infty }{\infty }}.$ Then,   $\lim _{x\rightarrow \infty }{\frac {f(x)}{g(x)}}=\lim _{x\rightarrow \infty }{\frac {f'(x)}{g'(x)}}.$ Solution:

Step 1:
First, let  $L=\lim _{n\rightarrow \infty }\left({\frac {n-7}{n}}\right)^{1/n}.$ Then,
${\begin{array}{rcl}\ln L&=&\displaystyle {\ln \left(\lim _{n\rightarrow \infty }\left[\left({\frac {n-7}{n}}\right)^{1/n}\right]\right)}\\\\&=&\displaystyle {\lim _{n\rightarrow \infty }\ln \left[\left({\frac {n-7}{n}}\right)^{1/n}\right]}\\\\&=&\displaystyle {\lim _{n\rightarrow \infty }\left[{\frac {1}{n}}\cdot \ln \left({\frac {n-7}{n}}\right)\right]}\\\\&=&0\cdot \ln(1)\\\\&=&0.\end{array}}$ Thus,  $L=e^{0}=1.$ Step 2:
Therefore, the sequence converges.
Additionally, the limit of the sequence is $e^{0}=1.$ The sequence converges. The limit of the sequence is  $1.$ 