Proving a fraction involving infinity tends to 0

But $n - 2n = -n \to -\infty$, and $\frac{n}{2n}$ does not go to 0, so I also am not sure what you are hoping to do.

You can't directly compare the two like that, it's absurd. Look at what I am saying. I am trying to point out that this becomes:

3^y where y=ln(n)/ln(3) - n , and that y tends to negative infinity as n tends to infinity. And a constant to the power of a large negative will tend to zero.

Okay, I understand your approach now, but how are you planning on showing $\frac{\log n}{\log 3} - n \to -\infty$ ?

By showing that it is a decreasing function, and that it has no minimum value - using calculus.

Or, to put it another (somewhat uncharitable) way, the method you suggested in post #16 (which mentioned neither log n or calculus) was seriously incomplete.

To agree with GE: If you'd written what you did in #16 in an exam I was marking, you'd get 0.

More generally, seeing this can be proved using only basic algebra (as opposed to using log, exp, infinite series or calculus), I find most of the solutions suggested somewhat inelegant.

I'm sorry, was I filling in an exam paper, I wasn't informed of this.

A proof is a proof and that's that.

Elegance is one thing, but I don't remember being asked how this could be most elegantly proven, and my method relies on very elementary Mathematics.

Back down, this isn't a position from which those who disagreed with me can argue solidly any longer. My solution is an obvious and valid one.

Infinity is NOT a real number and you can't treat it like one.

My solution is an obvious and valid one.

And a constant to the power of a large negative will tend to zero.

Not true. Consider, $(1/3)^{n}$. As n tends to negative infinity, this this does not tend to 0.

Uh? Yet again: n - 2n tends to -infinity, but n/2n does not tend to 0. Your proof is simply not valid.

@v-zero: I understand why you are saying what you saying, and why you think it is sensible. However, anybody who has taken any course on limits (i.e. Analysis) would never think of taking the approach you are attempting. The question is an elementary one on limits, such as those encountered at the beginning of a course, but you are having to assume numerous results that would take hours more work to justify properly. David's proof uses one simple theorem, that of the comparison test, but your attempts require so many unstated theorems and lemmas that it cannot (even if made to work!) constitute a suitable proof given the level of the question.

I think it's just something to put down to experience...

However, are you seriously arguing that supposing the groundwork was in place, that the limit of one divided by a term tending to infinity is not zero?

Well, probably one of the first things you will prove at university is that that assumption is actually true!

Rather than pull in exp, ln, and infinite series, note that n < 2^n, so $0 < \frac{n}{3^n} < \frac{2^n}{3^n} = \left(\frac{2}{3}\right)^n$ and the RHS clearly tends to 0 as n goes to infinity.

lax. stop having such a rant at v-zero.

Define yourself properly Franklin... n/3^n is not always greater than 0!! Try n = -1!

The post you quoted wasn't even a response to v-zero.

Seeing as we're discussing what happens as $n \to \infty$ your comment is not terribly relevant.

Not relevant, no. But then again, nor were your comments about v-zero's solution, given that we weren't trying to write a formal proof that would pass the rigorous assessment of a university analysis professor.