Looking for a counterexample for this statement regarding divergent series

Question

I need counter for this statement :

If the series $ \sum a_n$ is divergent then the series $b_n$= $\sum \text{min}(a_n,\frac{1}{n})$ is also divergent.

I closest I reached to a counter is , Define $a_n$ as $$a_n = \begin{cases} \frac{1}{n²} \text{for n even} \\ \frac{1}{n} \text{(any divegrent series) for n odd} \end{cases} $$ Here, $\sum a_n$ is divergent. Because $\sum 1/n²$ is convergent but $\sum 1/n$ is not. So their sum is divergent.

Now, in $b_n$ At times when $n$ is even $a_n$ (is $\sum 1/n²$ will converge ) will be less than $1/n$ and could be a counter but at odd it will cause problem when because when $n$ is odd there will be $1/n$ equal to it and is divergent but i need it to converge.

If this statement is true then a proof is needed.

Kindly Help.

Thank You.

Answer 1

Consider the sequence $$ a_n = \begin{cases} 0 & \text{if $n$ is not a power of $2$}\\ 1 & \text{if $n$ is a power of $2$}. \end{cases} $$ Since there are infinitely many powers of $2$, $\sum a_n$ diverges. However, $$ b_n = \min\left\{a_n,\frac{1}{n}\right\} = \begin{cases} 0 & \text{if $n$ is not a power of $2$} \\ \frac{1}{n} & \text{if $n$ is a power of $2$} \end{cases} $$ satisfies that $\sum b_n$ converges. Indeed, $$ \sum_{n=0}^{\infty} b_n = \sum_{k=0}^{\infty} \frac{1}{2^k} = 2. $$

Answer 2

Just to mention that the statement in the OP holds if in addition is is assumed that $a_n\searrow0$. The proof of that can be obtained by Cauchy's condensation.

Notice the $a_n\searrow0$ implies that $\min(a_n,\frac1n)\searrow0$. If $\sum_n\min(a_n,\frac1n)<\infty$, $\sum_k2^k\min(a_{2^k},2^{-k})=\sum_k\min(2^ka_{2^k},1)<\infty$ and so, $\sum_k2^ka_{2^k}<\infty$ which in turn means that $\sum_na_n<\infty$.