Proving a fraction involving infinity tends to 0

Of course I'm not, but you didn't attempt to put that groundwork in place. I don't think you even went anywhere near taking a quotient. You stated that n > log n / log 3, and therefore 3^n > n. Perfectly true, and this gives us that n / 3^n < 1. We also know that, if n is positive, 0 < n / 3^n < 1, and it's perfectly possible to show by arm-waving that n / 3^n gets smaller as n gets bigger, because as n changes to n+1, the denominator increases by a factor of 3 and the numerator doesn't increase nearly that fast. I don't know whether you expected us to work all this out ourselves from your one inequality... however, none of this shows that it converges to zero. It might very easily tend to 0.000001 from above.


Yes, but it wasn't correct. This isn't pedantry, or 'prudishness'. Your method simply wasn't right, and SimonM has pointed out a perfectly good counterexample which you seem to refuse to accept. It might seem intuitively obvious that it's true, but the things you were claiming are simply false. I'm sorry if you feel we've been rather rude to you, but it remains true that what you wrote wasn't just "unrigorous", and we're not just being pretentious arses - what you wrote was logically completely incorrect, and it's important that this is pointed out, not necessarily for your benefit (though I'd have hoped you'd like to learn from your own mistakes and not just dismiss it as mindless criticism), but for the benefit of anyone else - like the OP - who might have an interest in this kind of stuff. And, of course, for the pure intellectual honesty of answering the question correctly rather than giving an intuitive, but incorrect, reasoning as to why the limit of that sequence is as claimed.

I still don't think you understand my method, and I still don't agree that it is invalid.

Then explain your method. I'm quite certain that I've explained why whatever you said was invalid.

I have to ask: are you planning to tell your university interviewer that "my method's right, you just don't understand it" as well?

The difference here is that if I piss you off, it doesn't matter

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.

In maths today, our teacher posed a question which involved a infinite sum ending in

$\frac{n}{3^n}$

Me and my friend instinctively saw this tends to 0 as $n \rightarrow \infty$, and our maths teacher accepted that it must, given that we assumed this and got the correct answer. However, he insists it cancels with another term (we don't understand how this can be, seeing as we must have assumed the term it cancelled with didn't cancel, yet still got the right answer).

But all that is largely irrelevant. The important bit is this: he challenged us to prove this, and we just can't. We've looked around, but everything we've seen makes the same assumption that we do, without proof. I tried reasoning that, since

$\frac{1}{\infty} = 0$

anything, even infinity, multiplied by this must equal 0, giving the odd result

$\frac{\infty}{\infty} = 0$

This can sort of be explained away with 'infinity is weird like that'. However, my friend pointed out that this would imply

$\lim_{n \to +\infty} \frac{n^3}{n^2} = 0$

when, quite clearly, it actually is n (infinity). So that line of reasoning obviously doesn't work.

I wondered whether it involved aleph numbers, but since the last time I encountered them when watching a talk a couple of years ago, I've no idea. So, can anyone offer a proof?

The proof is simple. Infinity is continuous and thus has no real value and thus can't be expressed as anything other than zero.

In maths today, our teacher posed a question which involved a infinite sum ending in

$\frac{n}{3^n}$

Me and my friend instinctively saw this tends to 0 as $n \rightarrow \infty$, and our maths teacher accepted that it must, given that we assumed this and got the correct answer. However, he insists it cancels with another term (we don't understand how this can be, seeing as we must have assumed the term it cancelled with didn't cancel, yet still got the right answer).

But all that is largely irrelevant. The important bit is this: he challenged us to prove this, and we just can't. We've looked around, but everything we've seen makes the same assumption that we do, without proof. I tried reasoning that, since

$\frac{1}{\infty} = 0$

anything, even infinity, multiplied by this must equal 0, giving the odd result

$\frac{\infty}{\infty} = 0$

This can sort of be explained away with 'infinity is weird like that'. However, my friend pointed out that this would imply

$\lim_{n \to +\infty} \frac{n^3}{n^2} = 0$

when, quite clearly, it actually is n (infinity). So that line of reasoning obviously doesn't work.

I wondered whether it involved aleph numbers, but since the last time I encountered them when watching a talk a couple of years ago, I've no idea. So, can anyone offer a proof?

Please don't resurrect 14 year old threads - especially if you're just trolling!