2
$\begingroup$

I am trying to prove $\displaystyle \lim_{n \to \infty} \frac{n^a}{c^n} = 0$ using L'Hôpital's Rule, but I'm stuck.

Here's what I have so far:

$$ \lim_{n \to \infty} \frac{n^a}{c^n} = \lim_{n \to \infty}\frac{an^{n-1}}{c^n \ln c} = \lim_{n \to \infty}\frac{a(a-1)n^{a-2}}{c^n(\ln c)^2 + c^n \frac{1}{c}}$$

All three limits above seem to evaluate to $\frac{\infty}{\infty}$, so I feel like I'm not getting anywhere. Any ideas?

Edit: So, with the help of the hints below, I was able to figure out that

$$ \lim_{n \to \infty} \frac{n^a}{c^n} = \frac{a}{\ln c} \cdot \lim_{n \to \infty} \frac{n^{a-1}}{c^n} = \frac{a}{\ln c} \cdot \frac{a - 1}{\ln c} \cdot \lim_{n \to \infty} \frac{n^{a-2}}{c^n} = \cdots $$

So, disregarding the constant, it looks like the numerator keeps decreasing, while the denominator stays the same.

I can also see that if I let $a = 2$, for instance, I end up with $0$ after applying L'Hopital's $2$ times:

$$ \begin{aligned} \lim_{n \to \infty} \frac{n^2}{c^n} &\overset{LH}= \lim_{n \to \infty} \frac{2n}{c^n \ln c} \\ &= \frac{2}{\ln c} \lim_{n \to \infty} \frac{n}{c^n} \\&\overset{LH}= \frac{2}{\ln c} \lim_{n \to \infty} \frac{1}{c^n \ln c} \\ &= \frac{2}{(\ln c)^2} \lim_{n \to \infty} \frac{1}{c^n} \\ &= 0 \end{aligned} $$

So it seems reasonable to conclude that for an arbitrary $a > 0$, I will end up with $0$ after applying L'Hopital's $a$ times.

But I'm not sure how to go about using induction to prove it formally. I've only proven very simple sums by induction so far. Do I have to apply it to a product here?

$\endgroup$
  • 1
    $\begingroup$ hint: let $k$ be an integer such as $k>a$. Then $\frac{n^a}{c^n} < \frac{n^k}{c^n}$. Now apply L'Hopital $k$ times. $\endgroup$ – Vasya Aug 9 at 20:12
  • 1
    $\begingroup$ Double-check that $1/c$ --- you're differentiating with respect to $n$, so $\ln c$ is a constant. $\endgroup$ – Neal Aug 9 at 20:18
2
$\begingroup$

I deleted my old answer, as it missed the point a bit (especially given the edits to the question). I'm going to expand on J.G.'s answer, since you seem to need a little extra help.

Let's prove $\lim_{n\to\infty} \frac{n^a}{c^n} = 0$, for $a \in \Bbb{R}$ and $c > 1$. (if $0 < c \le 1$, then the sequence does not tend to $0$, and for $c = 0$, the expression is undefined). We can tackle this in a number of cases, but the cases reduce back down to one case fairly easily, using the squeeze theorem.

Case 1: $a \in \Bbb{N}_0 = \{0, 1, 2, \ldots\}$, and $c > 1$
In this case, we use induction on $a$ (not $n$, as I originally suggested). When $a = 0$, then $$\frac{n^a}{c^n} = \frac{1}{c^n}.$$ This tends to $0$, a fact which you seem happy to assume. If you wished to prove it, observe that the sequence $a_n = \frac{1}{c^n}$ satisfies is decreasing, bounded below by $0$, and hence convergent. It also satisfies the recurrence relation $a_{n+1} = \frac{a_n}{c}$, so if $L$ is its limit, then taking the limit of both sides yields $L = \frac{L}{c} \implies (c - 1)L = 0$, and hence $L = 0$, as $c \neq 1$.

You probably could skip the above proof, but either way, the base case is established.

Now, suppose for some $k \in \Bbb{N}_0$ (and $c > 1$), we have $$\lim_{n \to \infty} \frac{n^k}{c^n} = 0.$$ Then, \begin{align*} \lim_{n \to \infty} \frac{n^{k+1}}{c^n} &= \lim_{n \to \infty} \frac{(k+1)n^k}{\ln c \cdot c^n} &\text{L'Hopital's rule} \\ &= \frac{k+1}{\ln c} \lim_{n \to \infty} \frac{n^k}{c^n} \\ &= \frac{k+1}{\ln c} \cdot 0 = 0 &\text{induction hypothesis.} \end{align*} By induction, we now have $\lim_{n \to \infty} \frac{n^a}{c^n} = 0$ for all $a \in \Bbb{N}_0$ and $c > 1$. That is, we have completed this case.

Case 2: $a \in \Bbb{R}$, and $c > 1$
To prove this case, simply choose any natural number $k$ such that $k \ge a$ (we can do this, due to the Archimedean property). Naturally, if we take a negative value of $a$, then just choose $k = 0$ (or $1$, or anything higher really). Then, note that for all $n$, $$0 \le \frac{n^a}{c^n} \le \frac{n^k}{c^n}.$$ The first case proved that $\frac{n^k}{c^n} \to 0$. Thus, by squeeze theorem, we have a proof for case 2.

We can even extend to $c < -1$ too!

Case 3: $a \in \Bbb{R}$, and $c < -1$
We prove this again by squeeze theorem. Note that, $$-\frac{n^a}{|c|^n} \le 0 \le \frac{n^a}{|c|^n},$$ and by case 2, both bounds tend to $0$, proving case 3.

Hope that helps, and sorry for the misleading hint.

$\endgroup$
3
$\begingroup$

Your second differentiation is wrong because you've tried differentiating the denominator with respect to $c$ instead of $n$. If we differentiate $a$ times for an integer $a\ge0$, our limit is $\lim_{n\to\infty}\frac{a!}{c^n\ln^a c}$, which $=0$ for $c>1$. (You can handle other values of $a$ by squeezing.)

$\endgroup$
1
$\begingroup$

Hint If $c>1$, the limit is trivial for $a \leq 0$. For $a>0$ show instead that $$\left( \lim\limits_{n \to \infty} \frac{n^a}{c^n} \right)^\frac{1}{a}=0$$

Then, raise both powers to $a$.

$\endgroup$

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.