Let $(a_n)_{n\geq 1}, (b_n)_{n\geq 1}$ be two sequence of positive real numbers such that $$ \lim_{n\to +\infty}\frac{a_1+a_2+\cdots+a_n}{n}=\lim_{n\to +\infty}\frac{b_1+b_2+\cdots+b_n}{n}=0.$$

Conjecture. For all $\epsilon>0$, there are infinitely many values of indices $k$ such that $a_k<\epsilon$ and $b_k<\epsilon.$

I think that this is true but I can not prove it now.

In the special case where $a_n = b_n$, that is, there is only one sequence, then one can argue easily using a contradiction argument. In the general case, the hard part is to show that the same set of indices are shared by both sequences $(a_n)_{n\geq 1}, (b_n)_{n\geq 1}$.


The sequence $(c_n)$ defined by $c_n = a_n + b_n$ satisfies $$ \lim_{n\to \infty}\frac{c_1+c_2+\cdots+c_n}{n}=0 $$ as well. It follows that $$ \liminf_{n \to \infty} c_n = 0 $$ and in particular, for every $\epsilon > 0$, $$ \max(a_k, b_k) \le c_k < \epsilon $$ for infinitely many $k$.

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Suppose by contradiction that exists $\varepsilon>0$ and $N \in \mathbb{N}$ such that $c_n:=a_n+b_n\geq\varepsilon$, for $n>N$. Then

$$ \frac{c_N+\dotsc+c_n}{n}\geq\frac{(n-N+1)}{n}\varepsilon \to \varepsilon, $$ which is a contradiction. Therefore there exists infinitely many $k$s such that $a_k,b_k<c_k<\varepsilon$.

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  • $\begingroup$ I think you misunderstood the question. Both $a_k\lt\epsilon$ and $b_k\lt\epsilon$ for the same ks. $\endgroup$ – joriki Jun 22 '18 at 7:49
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    $\begingroup$ No, this answer is correct. If $a_n > \epsilon $ or $b_n > \epsilon $ then $a_n +b_n> \epsilon $ $\endgroup$ – Kavi Rama Murthy Jun 22 '18 at 9:06
  • $\begingroup$ Strictly speaking, shouldn't the conclusion be "... such that $a_k,b_k \le c_k < \epsilon$"? Consider e.g. $a_n = b_n = 0$ for all $n$. $\endgroup$ – Neal Young Jun 22 '18 at 20:48
  • $\begingroup$ The sequences are of positive numbers. $\endgroup$ – Gonzalo Benavides Jun 22 '18 at 21:23


$$ F(n) = {\sum_{i=1}^n{a_i}\over n}\\ G(n) = {\sum_{i=1}^n{b_i}\over n} $$

The conditions above are that $$ \lim_{n\to\infty}F(n) = \lim_{n\to\infty}G(n) = 0 $$

Let $p(n,\epsilon)$ be the fraction of $a_i$ that are $\geq \epsilon$, for $i$ up to $n$. Likewise $q(n,\epsilon)$ be the fraction of $b_1 \geq \epsilon$ for $i \leq n$.

Derive $F'(n,\epsilon)$ as follows: round each $a_i$ down to zero if $a_i < \epsilon$ or to $\epsilon$ if $a_i \geq \epsilon$. Clearly, since each term in this sum is less than or equal to the sum in $F(n)$, it must always be the case that $$\forall \epsilon > 0: F'(n,\epsilon) \leq F(n)$$

Also, it is straightforward to calculate: $$ F'(n,\epsilon) = p(n,\epsilon) \times \epsilon $$

Thus, we can conclude that since the limit of $F(n)$ is zero, so must also be the limit of $F'(n,\epsilon)$, and since $\epsilon$ does not vary with $n$:

$$ \forall \epsilon: \lim_{n\to\infty}p(n,\epsilon) = 0$$

By the definition of limit (and the fact that $1/2 > 0$), there must exist an M such that for all $n > M$, $p(n,\epsilon) < 1/2$.

Likewise for $q(n,\epsilon)$; there must be an $M'$ such that for all $n > M'$, $q(n,\epsilon) < 1/2$.

The minimum possible overlap of cases where $a_i < \epsilon$ and $b_i < \epsilon$ is:

$$r(n,\epsilon) = 1 - p(n,\epsilon) - q(n,\epsilon)$$

Let $T = \max(M,M')$. Clearly, for all $n > T$: $r(n,\epsilon) > 0$. Thus, we have an infinite set of numbers of which a positive fraction must have $a_i < \epsilon$ and $b_i < \epsilon$. Thus, there must be an infinite number of such cases.

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Well, one way to think about it:

As $\lim \frac{\sum a_i}n \to 0$ and $\lim \frac{\sum b_i}n \to 0$ then $\lim (\frac{\sum a_i}n + \frac{\sum b_i}n) \to 0+ 0 = 0$.

And $(\frac{\sum a_i}n + (\frac{\sum a_i}n = \frac {\sum_{i=0}^n (a_i + b_i)}n$ so

$\lim_{n\to \infty}(\frac{\sum a_i}n) + \lim_{n\to \infty}(\frac{\sum a_i}n) = \lim_{n\to \infty}\frac {\sum_{i=0}^n (a_i + b_i)}n =0$.

(Note: We CAN'T do this with sums that do not converge.)

Now for $c_i = a_i + b_i$ for any $\epsilon > 0$, I claim there are infinite many $c_i < \epsilon$.


Because if there are were only finitely many then there would be a last $c_N < \epsilon$ (or no $c_i < \epsilon$ at all) all for all $k > N$ then $c_k \ge \epsilon$. If $\sum_{i=0}^N c_i = K$ then $\sum_{i=0}^{n; n > N} c_i = K + \sum_{i=N+1}^n c_i \ge K + (n-N)\epsilon$.

So $\frac {\sum_{i=0}^n c_i}n \ge \frac Kn + \frac (n-N){n}\epsilon = \frac Kn -\frac Nn\epsilon + \frac nn\epsilon = \frac {K-N\epsilon}n + \epsilon$.

So $\lim_{n\to \infty} \frac {\sum_{i=0}^n c_i}n \ge \lim_{n\to \infty}\frac {K-N\epsilon}n + \epsilon = \epsilon > 0$.

That's a contradiction so there are infinitely many $c_n < \epsilon.


And that mean there are infinitely many $a_k + b_k < \epsilon$ and as each $a_k, b_k$ is positive $a_k < a_k +b_k < \epsilon$ and $b_k < a_k + b_k < \epsilon$.

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