6

Suppose $a_n\downarrow 0, \sum\limits_{n=1}^{\infty}a_n=+\infty, b_n=\min\{a_n,1/n\}$. Prove that $\sum b_n $ diverges.

In fact, I have known that two positive divergent series $\sum a_n ~\sum b_n$, $c_n=\min\{a_n,b_n\}, \sum c_n$ is not always divergent. But I do not know why this above series is surely divergent. Sincerely thanks.

Faye Tao
  • 1,814

2 Answers2

2

Hint: Use Cauchy condensation test $\displaystyle\sum_{n=1}^{\infty}a_{n}$ convergent,if and only if $\sum_{n=1}^{\infty}2^na_{2^n}$ convergent

so we only show that $$\sum_{n=1}^{\infty}2^n\min\left(a_{2^n},\dfrac{1}{2^n}\right)$$ divergent

if some postive integer $n$ such $a_{2^n}\ge\dfrac{1}{2^n}$,then $$\sum_{n=1}^{\infty}2^n\min\left(a_{2^n},\dfrac{1}{2^n}\right)=\sum_{n=1}^{\infty}1$$ divergent

if some postive integer $n$ such $a_{2^n}<\dfrac{1}{2^n}$,then $$\sum_{n=1}^{\infty}2^n\min\left(a_{2^n},\dfrac{1}{2^n}\right)=\sum_{n=1}^{\infty}2^na_{2^n}$$ is divergent,because $\sum_{n=1}^{\infty}a_{n}$ divergent,then you can use Cauchy condensation test

math110
  • 93,304
  • why download? can you explain? – math110 Jun 22 '15 at 10:55
  • if $\sum_{n=1}^{\infty}a_{n},a_{n}>0$ and $\sum_{n=1}^{\infty}b_{n},b_{n}>0$ divergent, then $\sum {n=1}^{\infty}(a{n}+b_{n})$ divergent – math110 Jun 22 '15 at 11:03
  • 1
    you can part two set $A={a_{n}|a_{2^n}\le \dfrac{1}{2^n}},B={a_{n}|a_{2^n}>\dfrac{1}{2^n}}$,then you can consider $A\bigcup B$ – math110 Jun 22 '15 at 11:05
  • 1
    The condensation test only holds for non-increasing sequences. While $1/n$ is nonincreasing, we only know that $a_n$ approaches zero, and it is easy to construct a sequence (e.g., $a_n=0$ if $n$ is odd and $2/n$ otherwise) such that $\min(a_n,1/n)$ is not non-increasing. – Aaron Jun 22 '15 at 17:52
  • @Aaron: the notation $a_n \downarrow 0$ means that $a_n$ decreases to $0$. If this was your only objection, you might want to reconsider your vote. – Alex M. Jun 23 '15 at 19:30
  • @AlexM. I didn't vote. – Aaron Jun 23 '15 at 19:31
  • 3
    Even with the above clarification, @Aaron's question still stands: how do you know that $\min { a_n , \frac 1 n }$ decreases? (I don't suggest that it isn't, I only ask why it should be so.) – Alex M. Jun 23 '15 at 19:39
  • 1
    Indeed, you should add a line mentioning (i) that the test requires non-increasing sequence, and (ii) justifying that the sequence is indeed non-increasing (it is not hard, but it's necessary). – Clement C. Jun 30 '17 at 14:41
2

Suppose instead that $\displaystyle\sum_{n=1}^{\infty}b_n$ converges.$\;$ Then $\displaystyle b_n\downarrow0\;$ (since $a_n\downarrow 0$ and $\frac{1}{n}\downarrow0$),

so $\displaystyle\lim_{n\to\infty}nb_n=0$ (as in the links shown below);

so $b_n=a_n$ for $n\ge N$ (for some $N\in\mathbb{N}$) and therefore $\displaystyle\sum_{n=N}^{\infty}b_n=\displaystyle\sum_{n=N}^{\infty}a_n$ diverges.

This gives a contradiction, so $\displaystyle\sum_{n=1}^{\infty}b_n$ diverges.

user84413
  • 27,211
  • 1
    How do you obtain $\lim \limits _{n \to \infty} n b_n = 0$? If $a_n = \frac 1 {n+1}$, then $b_n = \min { \frac 1 {n+1}, \frac 1 n } = \frac 1 {n+1}$ so your conclusion seems not to hold. – Alex M. Jun 23 '15 at 19:33
  • 1
    @AlexM. I am using the result of http://math.stackexchange.com/questions/369669 – user84413 Jun 23 '15 at 19:46
  • 2
    @user84413 That should be including in the answer, not a comment (with justification as of why you can use the result; i.e., why $(b_n)_n$ is non-increasing). – Clement C. Jun 30 '17 at 14:43
  • See also https://math.stackexchange.com/questions/4603 – user84413 Jul 01 '17 at 00:01