if $lim_{x \to \infty} (\sum_{n = 1}^{x} a_{n} n) / x = 0$ then $\sum_{n = 1}^{\infty} a_{n}$ is convergent

By Brad Rodgers on Sunday, March 31, 2002 - 02:29 am:

Is it true that if

lim_{x \to \infty} (\sum_{n = 1}^{x} a_{n} n) / x = 0

then $\sum_{n = 1}^{\infty} a_{n}$ is convergent? It seems reasonable, and I can't seem to find a counterexample, but I can't prove it either...

Thanks,

Brad

By Michael Doré on Sunday, March 31, 2002 - 02:42 am:

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

By Michael Doré on Sunday, March 31, 2002 - 02:22 pm:

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.

By Brad Rodgers on Tuesday, April 02, 2002 - 05:03 am:

I see. Surely it is true that if

lim_{x \to \infty} (\sum_{n = 0}^{x} a_{n} n^{2}) / x = 0

then $\sum_{n = 0}^{\infty} a_{n}$ is convergent, right? Because we require $a_{n} < 1 / n^{2}$ , thus the series is less than $π^{2} / 6$ .

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

lim_{x \to \infty} (\sum_{n = 0}^{x} a_{n} f (n)) / x = 0

\sum_{n = 0}^{\infty} a_{n}

f (x)

x

Or does such a function exist?

Thanks,

Brad

Michael Doré on Tuesday, April 02, 2002 - 12:40 pm:

Another good question. I'm really not sure of the answer to your first question. It is not necessarily true that $a_{n} < 1 / n^{2}$ . All we know is that the long-run average of $n^{2} a_{n}$ is zero, but this doesn't even imply that $n^{2} a_{n}$ is bounded. I'll have to think about this.

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

$\sum_{r = 1}^{n} r^{k} a_{r} / n \to L$ as $n \to \infty$ (*)

Then:

$\sum_{r = 1}^{n - 1} r^{k} a_{r} / (n - 1) \to L$

And so:

$(1 - 1 / n) \sum_{r = 1}^{n - 1} r^{k} a_{r} / (n - 1) \to L$

$\sum_{r = 1}^{n - 1} r^{k} a_{r} / n \to L$

Subtracting from (*)

$n^{k} a_{n} / n \to 0$

and so $n^{k - 1} a_{n} \to 0$ . So $n^{k - 1} a_{n}$ is bounded so $| a_{n} | \leq A / n^{t}$ for some $t > 1$ and constant $A$ . Hence $\sum a_{i}$ is absolutely convergent.

However in your question, all we can get from this approach is that $n a_{n} \to 0$ which we already know isn't good enough to guarantee convergence ( $a_{n} = 1 / (n \ln n)$ ). 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?

By Michael Doré on Tuesday, April 02, 2002 - 02:36 pm:

Also if $a_{n} \geq 0$ for all $n$ then your conjecture is true. This is because:

$2 \times \sum_{r = 1}^{2 n} a_{r} r^{2} / (2 n) - \sum_{r = 1}^{n} a_{r} r^{2} / n \to 2 \times 0 - 0 = 0$

So $\sum_{r = n + 1}^{2 n} a_{r} r^{2} / n \to 0$ . Therefore this is bounded, so there exists and $A$ such that:

$\sum_{r = n + 1}^{2 n} a_{r} r^{2} / n < A$

for all $n$ . But $r^{2} > n^{2}$ so:

$\sum_{r = n + 1}^{2 n} a_{r} n^{2} / n < A$

and

$\sum_{r = n + 1}^{2 n} a_{r} < A / n$

So:

$\sum_{r = 2^{s - 1} + 1}^{2^{s}} a_{r} < A / 2^{s - 1}$

therefore:

$\sum_{r = 2}^{2^{s}} < A + A / 2 + \dots + A / 2^{s - 1} < 2 A$

and the convergence quickly follows. The challenge is when $a_{n}$ changes sign infinitely often...