Show that if $\sum_na_n=\infty$ and $a_n\downarrow 0$ then $\sum\limits_n\min(a_{n},\frac{1}{n})=\infty$

Let $a_n$ be a sequence decreasing to $0$, and $\sum {{a_n} = \infty }$.
Show that:$$\sum {\min \left( {{a_n},{1 \over n}} \right)} = \infty$$

If there's $N_0$ such that $\forall n>N_0: \min(a_n, {1\over n}) = {1\over n}$ or $\forall n>N_0: \min(a_n, {1\over n}) = a_n$ then, the problem is trivial.

Otherwise, let us define $b_n = \min(a_n, {1\over n})$ and two subsequences:
$$b_{n_k} = {1 \over n_k},\quad b_{n_l} = a_{n_l}$$

$$\sum {{b_n} = } \sum {{b_{{n_k}}}} + \sum {{b_{{n_l}}}} = \infty + \infty$$

Is that right?

• No, this is not right. Why must $\sum b_{n_k}=\infty$? Same for $\sum b_{n_l}$. – Andrés E. Caicedo Apr 4 '14 at 20:14
• OK. Actually I thought about it and it is wrong :) – AnnieOK Apr 4 '14 at 20:15
• Note you have not used that $a_n$ decreases to $0$. The result is false otherwise. – Andrés E. Caicedo Apr 4 '14 at 20:19
• – sdcvvc Apr 5 '14 at 9:07

There is a theorem (I believe called "Cauchy condensation test") that says that a series of positive decreasing terms $\sum_n s_n$ converges if and only if $\sum_k 2^k s_{2^k}$ converges. For your series you get that the series converges if and only if $\sum_k \min (2^k a_{2^k}, 1)$ converges. It is clear that this series converges if and only if $\sum_k 2^k a_{2^k}$ converges, which is if and only if $\sum_n a_n$ converges. So if $\sum_n a_n$ diverges then your series diverges.

• Thanks, but why is it $1$? at $min(2^ka_{2^k}, 1)$? – AnnieOK Apr 4 '14 at 20:42
• No, it should be a $1$, as written. It is not a minor issue, it is key to make the proof work. – Andrés E. Caicedo Apr 4 '14 at 20:50
• @AnnieOK I've rewritten one of the variables for clarity's sake (there was a namespace collision). Consider that $s_n$ is your $b_n$, i.e., $s_n = \min(a_n, \frac1n)$. Now, think about how you could write $2^ks_{2^k}$, given this definition of $s_n$. – Steven Stadnicki Apr 4 '14 at 20:55
• @AnnieOK The $2^k$ term in your sequence is $\min(a_{2^k},1/2^k)$. So if you multiply that by $2^k$ for the Cauchy condensation test you get $\min(2^k a_{2^k},2^k/2^k) = \min(2^k a_{2^k}, 1)$. – user2566092 Apr 4 '14 at 20:56
• @AnnieOK Do not forget to verify that the terms of this sequence are decreasing, so that you can indeed apply the test. – Andrés E. Caicedo Apr 4 '14 at 21:12

Consider the sequence $b_n:=\min(a_n,\frac{1}{n})$. Note that it is non-negative and non-increasing: $$0\leq b_{n+1}\leq a_n\quad\hbox{and}\quad 0\leq b_{n+1}\leq \frac1n.$$ Thus, one can apply the Cauchy condensation to prove the divergence of $\sum b_n$. Now, we want to show that $$\sum_{n=1}^\infty2^n\min\left(a_{2^n},\frac1{2^n}\right)=\infty.$$ Consider the following two sets: $$A=\{n\in{\bf N}:a_{2^n}\geq\frac{1}{2^n}\},\quad B:=\{n\in{\bf N}:a_{2^n}<\frac{1}{2^n}\}.$$ If $A$ is infinite, then $$\sum_{n=1}^\infty2^n\min\left(a_{2^n},\frac1{2^n}\right)\geq \sum_{n\in A}1=\infty.$$ If $A$ is finite, then there exists some positive integer $k$, such that $n\in B$ for all $n\geq k$. It follows that $$\sum_{n=1}^\infty2^n\min\left(a_{2^n},\frac1{2^n}\right)\geq \sum_{n=k}^\infty2^na_{2^n}=\infty$$ where for the last equality we use the Cauchy condensation test and the fact that $\sum_{n=k}a^n=\infty$.

• The accepted answer already mentions the Cauchy condensation test. Perhaps not as cleanly, though. – Pedro Tamaroff Jun 30 '17 at 15:59
• @PedroTamaroff: Indeed. It was unclear for me in the accepted answer that why "It is clear that this series converges if and only if $\sum_k 2^k a_{2^k}$ converges". I think it could be useful to write out the details how the theorem is applied. There is also another answer using the Cauchy condensation test for the same problem in an unclear (I would say logically wrong) way. – Jack Jun 30 '17 at 16:06