[Jonnty]

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



Me and my friend instinctively saw this tends to 0 as , 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



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



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



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?

[Swayum]

When you get an "infinity/infinity" situation (or a 0/0 situation), things become unclear and you'd have to have done an analysis course to understand what's really happening.

Regarding your original problem, n/3^n - it tends to 0 because 3^n increase faster than n. So the denominator grows a lot faster than the numerator, which means it must tend to 0.

[Jonnty]

(Original post by Swayum)
When you get an "infinity/infinity" situation (or a 0/0 situation), things become unclear and you'd have to have done an analysis course to understand what's really happening.

Regarding your original problem, n/3^n - it tends to 0 because 3^n increase faster than n. So the denominator grows a lot faster than the numerator, which means it must tend to 0.

I'm pretty sure we put that to him and he didn't agree, but I'll put it to him again with more confidence tomorrow. Thanks.

[Swayum]

If you know differentiation (C4 level), you can show 3^n grows faster than n if that helps.

Hint: write y = 3^n and take the natural logarithm.

If he still doesn't give, ask him to prove why the argument is fallicious.

[Kolya]

The first line of your post is confusing. Are you considering the sequence , or the sum ?

[tommm]

Well, you can note that

(where I've substituted for simplicity).

So your fraction becomes:

which clearly tends to 0 as n (or x) tends to infinity.

[DeanK2]

Did you know the integrate test for convergence fails for this problem (as it returns the fraction of the form x/3^x.

Is this common?

[benwellsday]

So you're saying the series 1/3 + 2/9 + 3/27 .... + n / 3^n is convergent to 0 as n -> infinity? Because it doesn't.
The sequence n / 3^n does converge to 0 though.
And as for your arguements with infinity, bad idea. What you have said is that the limit of n^3 / n^2 = infinity / infinity, but infinity doesn't generally work with the quotient rule like that. There are loads of examples, limit of n+1 / n = infinity / infinity = 0, which is clearly wrong.

[DeanK2]

(Original post by benwellsday)
So you're saying the series 1/3 + 2/9 + 3/27 .... + n / 3^n is convergent to 0 as n -> infinity? Because it doesn't.
The sequence n / 3^n does converge to 0 though.
And as for your arguements with infinity, bad idea. What you have said is that the limit of n^3 / n^2 = infinity / infinity, but infinity doesn't generally work with the quotient rule like that. There are loads of examples, limit of n+1 / n = infinity / infinity = 0, which is clearly wrong.

Wasn't the question about WHY there is convergence (and yes the OP has alot to learn about the quotients he is so recklesssly throwing about).

What do you think about the integrable test failing (is this a general set of quotients - i.e. n/(b^f(n)), where this test fails?

[DFranklin]

(Original post by DeanK2)
Did you know the integrate test for convergence fails for this problem (as it returns the fraction of the form x/3^x.

In what sense does it "fail"?

You are asking whether converges.

If a test gives you the information that it converges if converges, then it hasn't failed. You have reduced the problem from one dealing with sums to one dealing with a single term, which is all you can expect.

[tommm]

Guys, unless I'm mistaken, I think that the OP is only referring to the limit of , and not the series itself.

[DFranklin]

(Original post by tommm)
Well, you can note that

(where I've substituted for simplicity).

So your fraction becomes:

which clearly tends to 0 as n (or x) tends to infinity.

Rather than pull in exp, ln, and infinite series, note that n < 2^n, so and the RHS clearly tends to 0 as n goes to infinity.

[tommm]

(Original post by DFranklin)
Rather than pull in exp, ln, and infinite series, note that n < 2^n, so and the RHS clearly tends to 0 as n goes to infinity.

That's much better.

[johnferguson1324]

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

[SimonM]

(Original post by Swayum)
So the denominator grows a lot faster than the numerator, which means it must tend to 0.

You need to be more specific with things like that. For example -

[v-zero]

You need to show that log to the base 3 of n is less than n, which is easy, and hence there is your proof.

[generalebriety]

(Original post by v-zero)
You need to show that log to the base 3 of n is less than n, which is easy, and hence there is your proof.

It's not really as simple as showing "less than", though. See Simon's example above yours.

[v-zero]

(Original post by generalebriety)
It's not really as simple as showing "less than", though. See Simon's example above yours.

Ok, if you want to be a prude, you need to prove that that is the case - which is easy.

[generalebriety]

(Original post by v-zero)
Ok, if you want to be a prude, you need to prove that that is the case - which is easy.

I'm not really sure what you're getting at. n is less than 2n, but n/2n does not tend to 0. Similarly, n is less than 3^n, but that doesn't prove that n/3^n tends to 0. I'm also not really sure what your "prude" comment is about, but I do know that if you wrote "log₃n < n, so n/3^n --> 0" in a university exam you'd be laughed at. Unless, of course, you're using some clever property of exponentiation that I don't know about... but that wouldn't really help the OP (or his teacher) either.

[v-zero]

Basically, I want him to prove that log to the base 3 of n, minus n will tend to negative infinity, as n tends to infinity.