0
$\begingroup$

I have this question, Find the limit of the sequence $$a_n:= \frac{n^{2001}}{1.001^n}$$ as $n \to \infty$.

I presume that the limit is $0$ due to the exponential in the denominator, and also presume I am to use the squeeze theorem to show this, but I am finding it hard to find two bounds that tend to the same limit. Or do I need to use a different theorem?

We have not used logarithms to solve limits yet and this exercise is meant to be completed using theorems and rules such as squeeze theorem, ratio test, sum/product/quotient rules etc.

$\endgroup$
0
$\begingroup$

Ok, so ratio it is:

$$\frac{a_{n+1}}{a_n}=\frac{(n+1)^{2001}}{1.001^{n+1}}\cdot\frac{1.001^n}{n^{2001}}=\left(1+\frac1n\right)^{2001}\cdot\frac1{1.001}\xrightarrow[n\to\infty]{}\frac1{1.001}<1$$

and thus the sequence converges to zero.

$\endgroup$
2
$\begingroup$

You only need the upper bound, as $a_n\ge 0$.

Then, you can prove using induction that $$\frac{ n^{2001}}{1.001^n} \le \frac Cn $$for a certain $C$.

$\endgroup$
  • $\begingroup$ I was meaning finding a greater value and a lower value that both tend to the same limit so that, by the squeeze theorem, a_n tends to the same limit? $\endgroup$ – MathsUndergrad Nov 2 '14 at 12:02
  • $\begingroup$ yes, you can take $0\le a_n \le \frac Cn$. $\endgroup$ – mookid Nov 2 '14 at 12:06
0
$\begingroup$

You could take a logarithm: $$2001\ln n - n\ln 1.001=\Bigl(2001-\frac n{\ln n}\ln 1.001\Bigr)\ln n$$ This goes to $-\infty$ because $\dfrac n{\ln n}$ goes to $\infty$.

$\endgroup$
0
$\begingroup$

$\ln(a_n)=2001\ln(n)-n\cdot\ln(1.001)=\ln(n)\cdot(2001-\frac{n}{\ln(n)}\cdot1.001)$

the limit of $2001-1.001\frac{n}{\ln(n)}$ is $-\infty$ and $\infty \times (-\infty)= - \infty$ so the limit of $\ln(a_n)$ is $-\infty$

$\endgroup$
  • $\begingroup$ What about the $\ln n$ in front of the brackets,why can we neglect it? $\endgroup$ – kingW3 Nov 2 '14 at 11:57
  • $\begingroup$ limit of $ln(n)$ is $\infty$ and the limit of what is inside the brackets is $-\infty$ so the limit of the product is clearly $-\infty$ $\endgroup$ – Pierre Alvarez Nov 2 '14 at 11:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.