1

In Calculus 2 we just started on doing sequences and I understand that to find the limit you can use l'hopital's rule and the sandwich theorem and a few other tricks but I'm generally confused on when to use what.

So we have this problem: 3^n/n^3
and we need to state if it diverges or converges, and if convergent, find the limit.

I really have no idea how to go about solving this, if someone can point me in the right direction it would be much appreciated. Thanks.

nukenine
  • 685
  • Have you compared exponential functions to polynomials before? – mickep Mar 29 '15 at 18:27
  • I don't believe we've covered that yet – nukenine Mar 29 '15 at 18:29
  • 1
    See what happens when you for instance double $n$. The numerator is squared, the denominator is multiplied by $8$. Which of those operations do you think is strongest in the long run? – Arthur Mar 29 '15 at 18:29
  • The strongest would be the one that becomes largest? So I believe 3^n would in the long run be larger than n^3 – nukenine Mar 29 '15 at 18:32
  • yes so every time you double n you multiply the number by 9/8 , so it should not converge – avz2611 Mar 29 '15 at 18:39
  • If you covered Taylor expansions you could just see what happens if you take the logarithm (since $\log$ is continuous and strictly increasing). – A.P. Mar 29 '15 at 18:39
  • http://math.stackexchange.com/questions/1208683/the-convergence-of-an-alternating-series-test – Bumblebee Mar 29 '15 at 18:43
  • Oh ok so the way to look at this is that because the numerator will be so much larger than the denominator for large numbers, the whole sequence tends to go towards infinity? Therefore, it's divergent. – nukenine Mar 29 '15 at 18:48
  • @Jay That is the idea behind it, but that's all just intuition. A proof is required. – Git Gud Mar 29 '15 at 18:50

3 Answers3

2

$$3^n=(2+1)^n=2^n+2^{n-1}\cdot \binom{n}{1}+\cdots+2\binom{n}{n-1}+\binom{n}{n}$$

When $n\ge 4$, there's a term $\binom{n}{4}=\frac{n(n-1)(n-2)(n-3)}{4!}$, which is a polynomial of degree $4$ with a positive $n^4$ coefficient. All the other terms of this expansion are positive. Ratio of higher degree to lower degree polynomial with positive highest degree coefficients as $x\to\infty$ is $\infty$.

(you can see this by dividing both the numerator and denominator by $x^k$ ($k$ -- numerator's highest degree))

user26486
  • 11,331
1

You may observe that $$ f(x)=x \ln (3/2)-3 \ln x >0, \quad x>24, \tag1 $$ since $$ f'(x)=\ln (3/2)-\frac3x >0, \quad x>24. $$ Then you deduce from $(1)$ that $$ \frac{3^n}{n^3} > 2^n, \quad n>24, $$ and, as $n \to \infty$, $\dfrac{3^n}{n^3} \to+\infty$.

Olivier Oloa
  • 120,989
1

$$\frac{3^n}{n^3}= \left( \frac{\sqrt[6]{3}^n}{\sqrt{n}} \right)^6$$

Now, by Bernoulli inequality

$$\sqrt{3}^n \geq 1+(\sqrt[6]{3}-1)n>(\sqrt[6]{3}-1)n $$

Therefore $$\frac{3^n}{n^3}\geq \left((\sqrt[6]{3}-1)\sqrt{n} \right)^6=\left(\sqrt[6]{3}-1\right)^6 n^3$$

N. S.
  • 132,525