# Another series/sequence proof

1. Jun 17, 2009

### JG89

1. The problem statement, all variables and given/known data

Prove that if the infinite series a_1 + a_2 + ... + a_n, with positive decreasing terms, converges to a value A, then the limit as n approaches infinity of n*a_n = 0

2. Relevant equations

3. The attempt at a solution

Assume $$\forall n \in Z^{+} ,\exists \epsilon > 0: n(a_n) \ge \epsilon$$, where Z+ is the set of all positive integers and of course, epsilon is fixed. Now, if n(a_n) >= epsilon, then a_n >= epsilon/n

If epsilon >= 1, then a_n >= epsilon/n >= 1/n, for all n. But this can't be true because then
a_1 >= 1/1, a_2 >= 1/2, a_3 >= 1/3 and so on, implying that a_1 + a_2 + ... + a_n >= 1/1 + 1/2 + 1/3 + 1/4 + ... = harmonic series, which diverges, and so a_n diverges, but a_n really converges.

Now we consider 0 < epsilon < 1:

There exists a fixed positive number c such that c*epsilon = 1. If n(a_n) >= epsilon, then
cn(a_n) >= c*epsilon = 1, implying that c(a_n) >= 1/n for all positive integers n.

And so, c(a_1) >= 1/1, c(a_2) >= 1/2, c(a_3) >= 1/3 and so on, implying:

c(a_1) + c(a_2) + c(a_3) = c(a_1 + a_2 + ..) >= 1/1 + 1/2 + 1/3 + ... = harmonic series, which diverges, and so the infinite series c(a_1) + c(a_2) + .. must also diverge. However, c(a_1 + a_2 + ...) converges to the value cA, since c is a fixed value.

These two contradictions both lead to the conclusion that there doesn't exist an epsilon such that n(a_n) >= epsilon for all positive integers n, meaning that n(a_n) < epsilon for all positive epsilon and for n large enough, meaning that the limit is 0.

QED

This proof looks fine to me, but the solution seems too simple. What is wrong with the proof?

Last edited: Jun 17, 2009
2. Jun 17, 2009

### xaos

it seems that should work. maybe clarify a bit why you can have n*a_n > epsilon. also if you feel like you can put all this into a single step by noticing epsilon is fixed, while sum(1/n) goes unbounded independently of epsilon. you might need to prove that sum(1/n) is unbounded.

if sum(a_n) converges it has a bound M say so that

M>sum(a_n)>epsilon*(sum(1/N) > M for suff large N. contradiction. so you can choose epsilon > M/sum(1/N) for N sufficiently large...[edit: for any N!]

3. Jun 17, 2009

### Dick

It is too simple. You didn't use that the sequence is monotone decreasing, which is an essential assumption. There error starts with "If n(a_n) >= epsilon". You seem to be assuming that is true for all integers. You can't assume that. The correct statement would be that there is an epsilon such that for any N>0, there is an n>N such that n(a_n)>epsilon. That is what n(a_n) does not converge to zero means. The correct proof does look a lot like the proof that the harmonic series diverges. Want to try it again? Here's a counterexample to think about. Define a_i=0 except where i=2^j for some integer j, then a_i=1/i. See? No monotone decreasing, no theorem.

Last edited: Jun 17, 2009
4. Jun 17, 2009

### JG89

Ok, I just looked up the proof of the harmonic series and tried to model this proof after that, trying to get an infinite number of 1/2's...

Assume that there exists an epsilon such that for any N > 0, there exists a positive integer n > N such that n(a_n) > epsilon. Let n1,n2,n3, ... be integers such that n1*2 < n2, 2*n2 < n3, 2*n3 < n4 and so on and for each ni, ni(a_ni) > epsilon.

Then the infinite sum a1 + a2 + a3 + ... + an1 + a(n1 + 1) + ... + an2 + ... can be grouped as follows:

(a1 + a2 + ... + an1) + (a(n+1) + ... + an2) + (a(n2+1) + ... + an3) + ...

For the first group, a1 > a2 > ... > an1 > epsilon/n1 and so a1 + a2 + ... + an1 > n1(an1) = epsilon.

For the second group, a(n1 +1) + a(n1 + 2) + ... + a(n2) > (n2 - n1)a(n2).

Since n2 > 2n1, then n2 - 2n1 > 0 <-> 2n2 - 2n1 > n2 <-> 2(n2 - n1)> n2 <-> n2-n1>n2/2

And so a(n1+1) + ... + a(n2) > (n2-n1)a(n2) > (n2/2)(epsilon/n2) = epsilon/2.

For the third group, a(n2 + 1) + ... + a(n3) > (n3 - n2)a(n3) > (n3/2)(epsilon/n3) = epsilon/2

And so (a1 + a2 + ... + an1) + (a(n1 +1) + ... + an2) + ... > epsilon + epsilon/2 + epsilon/2 + ... = epsilon(1 + 1/2 + 1/2 + 1/2).

The series 1 + 1/2 + 1/2 + ... diverges (this is used to prove the harmonic series diverges), and since epsilon is fixed, epsilon(1 + 1/2 + 1/2 + ...) also diverges, showing that a1 + a2 + a3 + ... diverges, even though it really converges. This is a contradiction and so n*an has a limit of 0.

5. Jun 17, 2009

### Dick

That's it. Congratulations. I haven't read every line of the proof, because I'm lazy, but I see all of the right ideas. If you believe it, I do.