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 $$a_n = \begin{array} \{\frac{1}{n²} \text{for n even} \\ \frac{1}{n} \text{( any divegrent series) for n odd} \end{array} $$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.
