009C Sample Midterm 3, Problem 1

Test if the following sequence ${a_{n}}$ converges or diverges. If it converges, also find the limit of the sequence.

$a_{n}=\left({\frac {n-7}{n}}\right)^{1/n}.$

Foundations:
This a common question, and is related to the fact that
$\lim _{x\rightarrow \infty }\left(1+{\frac {\alpha }{x}}\right)^{x}\ =\ e^{\alpha }.$
In such a limit, the argument $1+\alpha /x$ tends to one as $x$ gets large, while we are raising that argument to an increasing power. Neither one really "wins", so we end up with a finite limit that is neither zero nor infinity.
On the other hand, in the exam problem the argument $(n-7)/n$ is always smaller than one, but tends to one as $n$ gets large, while the exponent $1/n$ tends to zero. These do not disagree, so the limit should be one, but we need to prove it.
Any time you have a function raised to a function, we need to use natural log and take advantage of the log rule:
$\ln \left(a^{b}\right)=b\ln(a).$
For example, to find $\lim _{x\rightarrow \infty }\left(1-{\frac {1}{x}}\right)^{x}$, you could begin by saying: Let $L=\lim _{x\rightarrow \infty }\left(1-{\frac {1}{x}}\right)^{x}.$ Then
$\ln L=\ln \left[\lim _{x\rightarrow \infty }\left(1-{\frac {1}{x}}\right)^{x}\right]=\lim _{x\rightarrow \infty }\ln \left[\left(1-{\frac {1}{x}}\right)^{x}\right],$ where we are allowed to pass the log through the limit because natural log is continuous. But by log rules,
$\ln \left[\left(1-{\frac {1}{x}}\right)^{x}\right]=x\ln \left(1-{\frac {1}{x}}\right).$
Thus
${\begin{array}{rcl}\lim _{x\rightarrow \infty }\ln \left[\left(1-{\frac {1}{x}}\right)^{x}\right]&=&\lim _{x\rightarrow \infty }x\ln \left(1-{\frac {1}{x}}\right),\\&=&\lim _{x\rightarrow \infty }{\frac {\ln \left(1-{\frac {1}{x}}\right)}{\frac {1}{x}}}\\&{\overset {l'H}{=}}&\lim _{x\rightarrow \infty }{\frac {{\frac {x}{x-1}}\cdot \left(-{\frac {1}{x}}\right)'}{\left({\frac {1}{x}}\right)'}}\\&=&-1.\end{array}}$ Note that $\lim _{x\rightarrow \infty }{\frac {\ln \left(1-{\frac {1}{x}}\right)}{\frac {1}{x}}}={\frac {0}{0}},$  so we can apply l'Hôpital's rule. Finally, since $\ln L=-1,\,\,L={\frac {1}{e}}.$
Again, such a technique is not required for this particular problem, as the exponent tends to zero. But the technique is common enough on exams to justify providing an example.

Solution:
Following the procedure outlined in Foundations, let $L=\lim _{n\rightarrow \infty }\left({\frac {n-7}{n}}\right)^{1/n}.$ Then
${\begin{array}{rcl}\ln L&=&\ln \left(\lim _{n\rightarrow \infty }\left[\left({\frac {n-7}{n}}\right)^{1/n}\right]\right)\\&=&\lim _{n\rightarrow \infty }\ln \left[\left({\frac {n-7}{n}}\right)^{1/n}\right]\\&&\lim _{n\rightarrow \infty }\left[{\frac {1}{n}}\cdot \ln \left({\frac {n-7}{n}}\right)\right]\\&=&0\cdot \ln(1)\\&=&0.\end{array}}$
Thus, $L=e^{0}=1.$ Also, most teachers would require you to mention that natural log is continuous as justification for passing the limit through it.

Final Answer:
The limit of the sequence is $e^{0}=1.$

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