Is it true that if

then

¥ å n=1

an

is convergent? It seems reasonable, and I can't seem to find a counterexample, but I can't prove it either...

Thanks,

Brad

I think a_n = 1/(n ln n) (n > 1) is a counter-example - but I'll check this in the morning. Interesting question.

Indeed it is well known (and easy to prove) that if a_n -> L as n-> infinity then (a₁ + ... + a_n )/n -> L. (This result is the basis for the Cesaro limit. The Cesaro limit of a sequence is equal to the limit of the average of the first n terms. The above result tells us that if the sequence has a limit then it also has a Cesaro limit, and the Cesaro limit is equal in value to the normal limit. The converse isn't true - the sequence 1,0,1,0,1,0,... doesn't have a normal limit but has a Cesaro limit of 1/2.)

Anyway, this immediately shows that a_n = 1/(n ln n) satisfies your first condition since n a_n -> 0. (Note you have to redefine a₁ but it really doesn't matter what you define it as.)

To show a₁ + a₂ + ... diverges you can use exactly the same technique used to show the harmonic series diverges. Obtain a lower bound for the sum between n = 2^N-1 + 1 and n = 2^N . You should find this lower bound is proportional to 1/N and the result follows by comparison with the harmonic series.

I see. Surely it is true that if

then

¥ å n=0

an

is convergent, right? Because we require a_n < 1/n², thus the series is less than p²/6.

What is the minimal function like this? e.g. what monotonically increasing function f(x) satisfies

(a)


lim x®¥

æ è

x å n=0

an f(n)

ö ø

/x=0


implies that


¥ å n=0

an


is convergent.
(b) f(x) is less than any other function satisfying (a) at a given x?

Another good question. I'm really not sure of the answer to your first question. It is not necessarily true that a_n < 1/n². All we know is that the long-run average of n² a_n is zero, but this doesn't even imply that n² a_n is bounded. I'll have to think about this.

Note if you had replaced n² by any higher power of n, the answer would certainly be yes. In fact if k > 2 all you need is for 1^k a₁+¼+ n^k a_n)/n to converge (not necessarily to zoro) and this guarantees a₁+¼ converges. To see this, note that if:

n å r=1

rk ar/n® L

as n®¥ (*)

Then:

n-1 å r=1

rk ar/(n-1)® L

And so:

(1-1/n)

n-1 å r=1

rk ar/(n-1)® L

n-1 å r=1

rk ar/n® L

Subtracting from (*)

n^k a_n/n® 0

and so n^k-1 a_n® 0. So n^k-1a_n is bounded so |a_n| £ A/n^t for some t > 1 and constant A. Hence

å

ai

is absolutely convergent.

However in your question, all we can get from this approach is that n a_n® 0 which we already know isn't good enough to guarantee convergence (a_n=1/(nlnn)). I suspect your conjecture is still true, but proving this has me stumped.

Your last question is easier - the answer is no. If f(x) satisfies (a) then so does f(x)/2. However maybe you are defining minimal function differently?

Also if a_n ³ 0 for all n then your conjecture is true. This is because:

2×

2n å r=1

ar r2/(2n)-

n å r=1

ar r2/n® 2×0 - 0 = 0

So

2n å r=n+1

ar r2/n® 0

. Therefore this is bounded, so there exists and A such that:

2n å r=n+1

ar r2/n < A

for all n. But r² > n² so:

2n å r=n+1

ar n2/n < A

and

2n å r=n+1

ar < A/n

So:

2s å r=2s-1+1

ar < A/2s-1

therefore:

2s å r=2

< A+A/2+¼+A/2s-1 < 2A

and the convergence quickly follows. The challenge is when a_n changes sign infinitely often...