A kind of Cauchy's Condensation Test?

Question

Let $f:[0,\infty) \to \mathbb{R}$ be a positive , monotone decreasing function and $ f(x) \to 0$ as $x\to \infty$

Let $\{a_n\}$ be a bounded sequence such that $a_n \ge 0$ and $\displaystyle\sum _{n=1}^{\infty} a_n=\infty$ .

Then prove that $\sum_{n\ge 1} f(n)$ and $\sum_{n\ge 1}a_nf(a_1+a_2+\cdots+a_n)$ converge or diverge together.

My attempt

The zero terms of $\{a_n\}$ contribute nothing to the convergence/divergence of

$$\sum_{n=1}^{\infty}a_n$$

and $$\sum_{n\ge 1}a_nf(a_1+a_2+\cdots+a_n).$$

So without loss of generality , let us assume that $a_n \gt 0, \forall n$. Since $a_n$ is bounded, we must have

$m_*\le a_n \le M^*$ , for some $m_*, M^*\gt 0$

So, we have

$$a_1+a_2+\cdots+a_n \ge nm_*$$

If $m_* \ge 1$, then

$$a_1+a_2+\cdots+a_n \ge n$$

$\Rightarrow f(a_1+a_2+\cdots+a_n) \le f(n)$, since $f(x)$ is monotone decreasing function.

So by Comparision test if $\sum f(n)$ converges, then $\sum_{n\ge 1}f(\sum _{i=1}^{n}a_i)$ converges and so does $\sum_{n\ge 1}a_nf(\sum _{i=1}^{n}a_i)$ since

$$a_nf\left(\sum _{i=1}^{n}a_i\right) \le M^*f\left(\sum _{i=1}^{n}a_i\right).$$

The case for $m_* \ge 1$ ends here.

Let $M^* \le 1.$

Then $a_1+a_2+\cdots+a_n \le nM^*\le n$

$$\Rightarrow f\left(\sum _{i=1}^{n}a_i\right)\ge f(n)$$

Then by Comparision Test, if $\sum_{n\ge 1}f(n)$ diverges , then $\sum_{n\ge 1}f(\sum _{i=1}^{n}a_i)$ diverges and so does $\sum_{n\ge 1}a_nf(\sum _{i=1}^{n}a_i)$ since

$$a_nf\left(\sum _{i=1}^{n}a_i\right) \ge m_*f\left(\sum _{i=1}^{n}a_i\right).$$

The case for $M^* \le 1$ ends here.

I do not know how to approach for $m_* \le 1 \le M^*$ and moreover I am familiar with the missing reverse implications in the above cases.

Do you have any ideas? Thanks for your time.

As a sidenote, this is a previously asked question on MSE but the question was voluntarily removed by the OP for some unknown reasons (at least unknown to me). I would really like the question to be answered.

Answer 1

PARTIAL answer:

$(a_n)$ is bounded so there exists an $M > 0$ such that for all $n$ we have $0 \leq a_n \leq M$. Moreover, as you suggested, we can assume without loss of generality that $a_n > 0$ for all $n$.

Let $N_k = \min \{ n : a_1 + \dots + a_n \geq k\}$.

$N_k$ is well defined and finite for all $k$ because $\sum_n a_n = \infty$.

Now since all terms are non-negative we can write

$$ \sum_{n \geq 1} a_n f(a_1 + \dots + a_n) = \sum_{k\geq 0} \sum_{n=N_k}^{N_{k+1}-1} a_n \underbrace{f(a_1 + \dots + a_n)}_{\leq f(k)} \leq \sum_{k \geq 0} \underbrace{\left(\sum_{n=N_k}^{N_{k+1}-1} a_n\right)} _{\leq a_{N_k} + 1 \leq M + 1} f(k) \leq (M + 1) \sum_{k \geq 0} f(k) $$

where the sum in parenthesis in the center may be empty and thus equal to zero.

This proves that the divergence of $\sum_n a_n f(a_1 + \dots + a_n)$ implies the divergence of $\sum_n f(n)$.

Now we write similarly

$$ \begin{aligned} \sum_{n \geq 1} a_n f(a_1 + \dots + a_n) &= \sum_{k \geq 0} \sum_{n=N_k}^{N_{k+1}-1} a_n \underbrace{f(a_1 + \dots + a_n)}_{\geq f(k+1)} \\ &\geq \sum_{k \geq 0} \left(\sum_{n=N_k}^{N_{k+1}-1} a_n\right) f(k+1) \\ &= \sum_{k \geq 1} \underbrace{\left(\sum_{n=N_{k-1}}^{N_k-1} a_n\right)}_{b_k} f(k) = \sum_{k \geq 1} b_k f(k). \end{aligned} $$

Notice that $ 0 \leq b_k \leq M + 1$ for all $k$, and that $\sum_k b_k = \sum_n a_n = \infty$.

In fact, we even have that for all $K$, $$ \max_{\substack{0 \leq k < K \\ N_k < N_K}} k \ \leq \sum_{k=1}^K b_k = \sum_{n=1}^{N_K-1} a_n \leq K $$

The idea here is that since $f$ is decreasing, you can always 'shift' the weights towards bigger values of $k$ to get a lower bound of $\sum_k b_k f(k)$. If you get a lower bound of the form $ C \sum_{k \geq N} f(k)$ with $N$ an integer and $C$ a constant you are done because this means that the divergence of $\sum_n f(n)$ implies the divergence of $\sum_n a_n f(a_1 + \dots + a_n)$. I believe that finding such a lower bound is doable.

There might be a nicer solution though...

Hope this helps a bit even though it is not a full solution!

Answer 2

I think I have found a solution. Here it goes .

Proof :-

Since $f$ is monotone , it is integrable on every bounded interval.

As I already said, let $a_n \gt 0$

Let $M=\sup \{a_n : n\in \mathbb{N} \}$

Let $s_n=a_1+a_2+..+a_n, n\in \mathbb{N} $

Therefore $s_n \to \infty$ as $n\to \infty$

Consider the partition $P$ of $[0, s_n]$ given by

$\{0, s_1, s_2,..., s_n\}$

Since $f(x)$ is decreasing , the lower sum

$L(P,f)=a_1f(a_1)+a_2f(a_1+a_2)+...+a_nf(a_1+a_2+...a_n)$.

$=\sum_{i=1}^n a_i f (s_i)$

So $L(P,f)$ is one of our required sum here.

We know that

$L(P,f) \le \int_0^{s_n}f(x)dx \quad (1)$

Similarly $U(P,f)$ is the Upper Reimann Sum and it can be easily proved

$U(P,f)-L(P,f) \le ||P|| \{f(0)-f(s_n)\}$

$\le Mf(0)=k$ (say), a constant

$L(P,f) \ge U(P,f)-k \ge \int_0^{s_n} f(x)dx-k $

Therefore from $(1)$

$\int_0^{s_n}f(x)dx -k \le L(P,f) \le \int_0^{s_n} f(x)dx $

and thus

$\int_0^{s_n}f(x)dx -k \le\sum_{i=1}^n a_i f (s_i) \le \int_0^{s_n} f(x)dx \quad (2) $

Now, let $F(x)=\int_0^x f(t) dt$.

Then $F(x)$ is monotone increasing since $f$ is positive.

Let $\lim _{n\to \infty} \sum_{i=1}^n a_i f (s_i)$ exist .

Then clearly, $\lim_{n\to \infty} F(s_n)$ exist from $(2)$. Now we must have

$\lim_{x\to \infty} F(x)\lt \infty$ otherwise $\lim_{x\to \infty} F(x)=\infty$ and hence $\lim_{n\to \infty} F(s_n)=\infty$, a contradiction.

So $\lim_{x\to \infty}F(x) \lt \infty \Rightarrow \lim_{n\to \infty} F(n)\lt \infty$

and hence $\sum_{n\ge 1} f(n)$ exist by Abels Integral Test

Thus if $\sum_{n\ge 1}a_nf(\sum_{i=1}^na_i)$ exist then $\sum_{n\ge 1}f(n)$ also exist.

Arguing in the reverse way, the converse can be shown.