Proving a fraction involving infinity tends to 0

Sorry, but what v-zero posted was pretty much nonsense.

Sorry, but what v-zero posted was pretty much nonsense. Whereas you would be very unlucky to lose a single mark for leaving out "$\forall n \ge 0$" in what I posted. For example, it's the kind of thing that a "university analysis professor" wouldn't bat an eyelid over.

Having hung around D&D too much, I've heard some bizarre comparisons in my time on TSR, but that must be the most puzzling response that I have seen on TSR. You're taking a rejection of a silly objection to a proof (obviously $n \in \mathbb{N}$ because it was a term in a series?), and comparing it to racism on the London Underground?

Isn't it excellent that our right to draw ridiculous comparisons is not limited by any requirement of proper definition or rigour?

As for the Maths, surely it is plain that in this case if you can prove it for $n \in \mathbb{R}$ it is proven for $n \in \mathbb{N}$ ?

Another piece of ******** that v-zero has come out with it. This question requires natural numbers (or was it the postive real numbers? (well one of the many questions asked do - which one - it is hard to keep track of))). Yet another rpiece of annoying smuggness that you believe you can pedanitcally argue your 'superior' and petty false tricks in order to "weaken" the arguments of the only people who have actually tried to help you. Get a life you dickhead.

Another piece of ******** that v-zero has come out with it. This question requires natural numbers (or was it the postive real numbers? (well one of the many questions asked do - which one - it is hard to keep track of))). Yet another rpiece of annoying smuggness that you believe you can pedanitcally argue your 'superior' and petty false tricks in order to "weaken" the arguments of the only people who have actually tried to help you. Get a life you dickhead.

Un+1= aUn +b.
next term after Un is n+1/3^n+1
this makes:
Un+1= Un((n+1)/3n)

L= L((n+1)/3n)
L(1-((n+1)/3n))=0
for (n+1)/3n:
both n+1 and 3n have a limit of infinity, as is seen by examining them in recurrence relation format:
in n+1:
a is one, so will not influence terms but b, equal to one, a constant positive is being added to each term to make the next one, increasing the value of each successive term by an equal amount ,
in 3n:
, a is 1, so will not influence terms but b is 3, a constant postive is being added to each term to make the next one,increasing the value of each successive term by an equal amount. .
so the fraction, at limit, becomes infinity/infinity=0
this means :
L(1-0)=0
so L=0

I couldn't see much of an argument for all the whining.

I don't lack respect for these people, in several past posts I have noted both their Mathematical superiority, and their greater knowledge - and when they have posted something particularly pleasing and I have seen it, I have often commented as such.
Tried to help me? Oh no no... They have done little or nothing of the sort.

Of what "petty false tricks" you speak I do not know.

If you prove n is an element of R you cannot assume that n is an element of the natural numbers.

i can prove 2.5 is an element of R. I know N is a subset of R. However, 2.5 certainly does not belong in the natural numbers. (Infinite many counterexamples).

In general I do not disagree, however in the context of the question it is enough to prove it for all real numbers.

Yes - so why did you assume n is an element of the natural numbers (I am just assuming that for the question you are talking about this is valid - I actually have lost count as to how many people have decidided to post up their own little sequence / question) etc.

Because he was talking about a series summation, and it seemed (to me) implicit. I actually assumed they were members of the integers (but that's a moot and quasi-pedantic point), and hence a solution that involved all real numbers was equally valid.

The question was how to prove that the limit of $\frac{n}{3^n}$ equals zero as n tends to positive infinity.

That is clearly not a summation question - I could put n into any class (bar the obvious examples - complex, negaitive R, etc), and still derive this result.

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?

Lastly I want to make something clear: I was posting a plausible and understandable method, not a proof, and that was always my intention.