Limit of sequence, squeeze theorem?

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.

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.

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 $C$.

• 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? – MathsUndergrad Nov 2 '14 at 12:02
• yes, you can take $0\le a_n \le \frac Cn$. – mookid Nov 2 '14 at 12:06

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 $\infty$.

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

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 $-\infty$

• What about the $\ln n$ in front of the brackets,why can we neglect it? – kingW3 Nov 2 '14 at 11:57
• 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 $-\infty$ – Pierre Alvarez Nov 2 '14 at 11:59