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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.

Gajjze
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  • It’s hard to understand what your question is – Andrew Feb 03 '24 at 17:07
  • Okay I'm editing it to make it more clear – Gajjze Feb 03 '24 at 17:09
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    Why do you think it’s false? It seems likely to be true, to me at least – Andrew Feb 03 '24 at 17:29
  • This is a statement from a MCQ ,It also seemed true to me but this was incorrect (I too don't know how) ,now need a proof for it. – Gajjze Feb 03 '24 at 17:35
  • I see, yes you are right, the statement is false – Andrew Feb 03 '24 at 17:42
  • It is true if $(a_n)$ is decreasing: https://math.stackexchange.com/q/739975/42969 – Martin R Feb 03 '24 at 17:46
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    Okay, the thin is that if have "bad" $a_n$ we need to have "good" $\frac 1n < a_n$. You have the idean that $\frac 1n^2$ are "good" so lets say If $n$ is a perfect square then $a_n$ is some term of a divergent sequence and $a_n = \frac 1{n^2}$ when $n$ is not a perfect square. Then $\sum b_k = \sum_{k\text{ not a square}}\frac 1{k^2}+\sum \frac 1{k^2}$. – fleablood Feb 03 '24 at 19:50
  • @fleablood that's a good approach for it, thank you sir. – Gajjze Feb 04 '24 at 01:24

2 Answers2

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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, $$ \sum_{n=0}^{\infty} b_n = \sum_{k=0}^{\infty} \frac{1}{2^k} = 2. $$

Didier
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  • Note the reindexing in the last line. The $b_n$ series represents all terms of the series, but the sum with k represents only the non zero terms of b. Of course dropping out the zero terms has no effect on the sum. n vs k took me a minute or so to realize. – nickalh Feb 04 '24 at 13:42
  • @nickalh Using proper English, the sum of all $b_n$'s is by definition the sum of all the inverse of the powers of $2$. No need to abstractly re-index. This is why I used words and not abstract notations to define $(a_n)$ and express $(b_n)$ – Didier Feb 04 '24 at 13:45
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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 $\sum_na_n<\infty$.

Mittens
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