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?
Proving a fraction involving infinity tends to 0
Announcements  Posted on  

Study Help needs new mods!  14042014  
Post on TSR and win a prize! Find out more...  10042014 


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. 
(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. 
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. 
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? 
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. 
(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.
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? 
(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.
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. 
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(10)=0
so L=0 
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.

(Original post by vzero)
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. 
(Original post by generalebriety)
It's not really as simple as showing "less than", though. See Simon's example above yours. 
(Original post by vzero)
Ok, if you want to be a prude, you need to prove that that is the case  which is easy. 
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.
Reply
Submit reply
Register
Thanks for posting! You just need to create an account in order to submit the post Already a member? Sign in
Oops, something wasn't right
please check the following: