# Thread: Simple question on proof formalism

1. ## Simple question on proof formalism

For this particular proof I'm writing I'm relying on the fact that, for all $\displaystyle n \in N$, $\displaystyle n > 0$, the sequence $\displaystyle (a_n) = 2n^n$ is divergent. Of course, the result is intuitive and obvious, but in the problem statement I've been asked to prove separately any supporting claims such as this one. So, given that the claim is quite trivial, would it be sufficient just to proceed along the lines of:

"For all $\displaystyle n \in N$, $\displaystyle n > 0$, we have $\displaystyle 2n^n > n^n > n$. Thus, taking $\displaystyle n = \infty$, we have

$\displaystyle (a_n) = 2n^n > \infty$

Hence $\displaystyle (a_n)$ is divergent."

Or should I set about trying to prove this more rigorously using the definition of a limit of a sequence?

2. Your basic idea works, but you want to clean up your notation. Mathematically, you're using the plain ol' comparison test: comparing $\displaystyle 2n^{n}$ with the sequence $\displaystyle n$. Once you've shown that $\displaystyle 2n^{n}>n$ for all $\displaystyle n>0$, then all you have to do is show that $\displaystyle (n)$ diverges. At that point, you invoke the comparison test, and you're done.

In particular, the statement $\displaystyle 2n^{n}>\infty$ is a bit sloppy. In addition, you can't write $\displaystyle n=\infty$; you can only take limits as $\displaystyle n\to\infty$.

Make sense?

3. Originally Posted by Ackbeet
Your basic idea works, but you want to clean up your notation. Mathematically, you're using the plain ol' comparison test: comparing $\displaystyle 2n^{n}$ with the sequence $\displaystyle n$. Once you've shown that $\displaystyle 2n^{n}>n$ for all $\displaystyle n>0$, then all you have to do is show that $\displaystyle (n)$ diverges. At that point, you invoke the comparison test, and you're done.

In particular, the statement $\displaystyle 2n^{n}>\infty$ is a bit sloppy. In addition, you can't write $\displaystyle n=\infty$; you can only take limits as $\displaystyle n\to\infty$.

Make sense?
Sure does! Thanks for clearing that up.

4. I would try to find $\displaystyle \lim_{n \to \infty}2n^n$

$\displaystyle \lim_{n \to \infty}2n^n = \lim_{n \to \infty}e^{\ln{2n^n}}$

$\displaystyle = e^{\lim_{n \to \infty}\ln{2n^n}}$

$\displaystyle = e^{\lim_{n \to \infty}(n\ln{2n})}$

$\displaystyle \to e^{\infty}$

$\displaystyle \to \infty$.

So the sequence is divergent.

5. You're welcome. Have a good one!

6. Ok guys, I've reached another dead end. To complete this proof I'll also need to show that the similar sequence $\displaystyle (a_n) = 1^n - 2n^n$ is also divergent. I don't think I can use the comparison test here, seeing as this sequence is $\displaystyle < 0$ for all n. I've also tried proving divergence directly from the definition (or by contradiction), but I'm not sure how to manipulate the inequality $\displaystyle |1^n - 2n^n - L| < \epsilon$ in order to show this. Any pointers?

7. Well, you can simplify things a bit there. $\displaystyle (a_{n})=1-2n^{n}.$

I'm not sure where this theorem is, but off-hand, I think it's true. If you can prove that an arbitrary sequence $\displaystyle (b_{n})$ diverges if and only if $\displaystyle (-b_{n})$ diverges, you could then work with $\displaystyle (-a_{n})=2n^{n}-1;$ you could essentially use the result in the previous problem.