# Thread: lim as n approach infinity

1. ## lim as n approach infinity

Hello.

I got stuck on this:

Show using the binomial theorem, that if $a>1$ then $\lim_{n\rightarrow\infty}a^n=+\infty$

Here´s my brave attempt:

$(1+(a-1))^n=\sum_{k=1}^{n}\binom{n}{k}1^{n-k}(a-1)^k$

$=\binom{n}{1}(a-1)+\binom{n}{2}(a-1)^2+...+\binom{n}{n}(a-1)^n$

So at first I thougt that each term is bigger than one (this is the part that I got wrong). Then, since there are n terms, the sum must tend to infinity because n tends to infinity and since each term bigger than one.

The b part:

Show that
$\lim_{n\rightarrow\infty}n^{-1}a^n=+\infty$

I want to try this myself after getting the a) part straight but I wonder is there a clue in the b part for solving a?

Any hints, besides replacing my toilet paper with my home assignment?

2. Originally Posted by wilbursmith
Hello.

I got stuck on this:

Show using the binomial theorem, that if $a>1$ then $\lim_{n\rightarrow\infty}a^n=+\infty$

Here´s my brave attempt:

$(1+(a-1))^n=\sum_{k=1}^{n}\binom{n}{k}1^{n-k}(a-1)^k$ Not quite right – the sum should start with k=0, not k=1.

$=\binom{n}{1}(a-1)+\binom{n}{2}(a-1)^2+...+\binom{n}{n}(a-1)^n$

So at first I thougt that each term is bigger than one (this is the part that I got wrong). Then, since there are n terms, the sum must tend to infinity because n tends to infinity and since each term bigger than one.
You have made a good start. So you have the equation $a^n=1+\binom{n}{1}(a-1)+\binom{n}{2}(a-1)^2+...+\binom{n}{n}(a-1)^n.$ You don't need each term on the right to be greater than 1. You just need to say that each term is positive, so that $a^n$ is greater than any single term on the right. In particular, $a^n > \binom{n}{1}(a-1)$, and that tends to infinity as n does.

Originally Posted by wilbursmith
The b part:

Show that
$\lim_{n\rightarrow\infty}n^{-1}a^n=+\infty$

I want to try this myself after getting the a) part straight but I wonder is there a clue in the b part for solving a?
Yes there is. This time, try using one of the other terms on the right-hand side of that binomial expansion.

3. Ok I think the a part is rather clear now. The only way that I see $a^n$ is bigger than each term though is that a sum is always bigger than any of it´s terms assuming all positive (not much proof there). Is there a more convincing way of seeing that $a^n$ actually is bigger than any of the terms?

To the b part. I thought that $\lim\frac{a^n}{n}\rightarrow\infty$ since numerator grows faster than denominator, but I quess my prof will repeat my name three times with decreasing pitch while shaking his head as he reads my conclusion.

I wanted to do a binomial expansion of the expression but can´t split the expression the same way as in part a

e.g. $a^n=(1+(a-1))^n$

Maybe I don´t need to do the binomial but use what I got i part a somehow?

4. Originally Posted by wilbursmith
Ok I think the a part is rather clear now. The only way that I see $a^n$ is bigger than each term though is that a sum is always bigger than any of it´s terms assuming all positive (not much proof there). Is there a more convincing way of seeing that $a^n$ actually is bigger than any of the terms?
A sum of positive terms is always greater than any of its individual terms. That should be convincing enough for anyone.

Originally Posted by wilbursmith
To the b part. I thought that $\lim\frac{a^n}{n}\rightarrow\infty$ since numerator grows faster than denominator, but I quess my prof will repeat my name three times with decreasing pitch while shaking his head as he reads my conclusion.

I wanted to do a binomial expansion of the expression but can´t split the expression the same way as in part a

e.g. $a^n=(1+(a-1))^n$

Maybe I don´t need to do the binomial but use what I got i part a somehow?
Use the same binomial expansion as in part (a). But instead of saying that $a^n$ is greater than the term ${n\choose1}(a-1)$, say that $a^n$ is greater than the next term ${n\choose2}(a-1)^2$.

5. (b) Show that $\lim\frac{a^n}{n}\rightarrow\infty$

Ok, since $a^n >$ than any of the terms as $n\rightarrow\infty$ then

$a^n>\binom{n}{2}(a-1)^2$. Divide by (positive) n

$\frac{a^n}{n}>\binom{n}{2}\frac{(a-1)^2}{n}=\frac{n!(a-1)^2}{2n(n-2)!}=\frac{n(n-1)(n-2)...(a-1)^2}{2n(n-2)(n-3)...}=\frac{(n-1)(a-1)^2}{2}\rightarrow\infty$ as $n\rightarrow\infty$

6. Originally Posted by wilbursmith
(b) Show that $\lim\frac{a^n}{n}\rightarrow\infty$

Ok, since $a^n >$ than any of the terms as $n\rightarrow\infty$ then

$a^n>\binom{n}{2}(a-1)^2$. Divide by (positive) n

$\frac{a^n}{n}>\binom{n}{2}\frac{(a-1)^2}{n}=\frac{n!(a-1)^2}{2n(n-2)!}=\frac{n(n-1)(n-2)...(a-1)^2}{2n(n-2)(n-3)...}=\frac{(n-1)(a-1)^2}{2}\rightarrow\infty$ as $n\rightarrow\infty$

7. The assignment has a (c) part as well. Hope to get some help on that one too.

c) Show that for every $k\in N$

$\lim_{n\rightarrow\infty}n^ka^{-n}=0$

Earlier I stated that $a^n>\binom{n}{2}(a-1)^2
$
and so $\frac{1}{a^n}<\frac{1}{\binom{n}{2}(a-1)^2}
$
multiply by $n$ we get

$\frac{n}{a^n}<\frac{n}{(a-1)^2}\cdot\frac{2(n-2)(n-3)...}{n(n-1)(n-2)(n-3)...}=\frac{2}{(a-1)^2(n-1)}\rightarrow0$ as $n\rightarrow\infty$

Is this ok? And if so, how do I show that this is true for every $k\in N$...induction? How?

Again it feels natural that if I keep multiplying both sides with (positive)n, k times I´ll still get left side less than right side which tends to zero as n tends to infinity.

hmm?