Looking to get some help with the following question.

I need to find nonnegative sequences $(a_n)$and $(b_n)$ such that $a_n\leq b_n$ for all $n$ and $\sum a_n$ converges but $\sum b_n$ diverges.

So the terms of $a_n$ must turn to $0$ but $b_n$ does not. But how do I find examples of this series?

  • $\begingroup$ @Arrow thank you for the edit $\endgroup$
    – mp12345
    Jan 18, 2017 at 23:13
  • $\begingroup$ @SimpleArt sorry I had a typo there, just made a edit $\endgroup$
    – mp12345
    Jan 18, 2017 at 23:15
  • $\begingroup$ But it's not a question. $\endgroup$
    – zhw.
    Jan 18, 2017 at 23:16
  • 3
    $\begingroup$ Why can't $b_{n}$ tend to zero? These terms can still constitute a divergent series. For example: $$ a_{n} = {1 \over n^2}, \quad b_{n} = {1 \over n}. $$ $\endgroup$
    – avs
    Jan 18, 2017 at 23:17

1 Answer 1


Well, if $a_n=\frac1{2^n}$ and $b_n=n$, then

$$a_n\le b_n$$


$$\sum_{n=1}^\infty n=+\infty$$

But it is not necessary for $\require{cancel}b_n\cancel\to0$, for example, with $b_n=\frac1n$,

$$\sum_{n=1}^\infty\frac1n>\int_1^\infty\frac1x\ dx=\lim_{t\to\infty}\ln(t)=+\infty$$

But it is necessary for $a_n\to0$ by the term test.

  • $\begingroup$ no I understand this now, thanks $\endgroup$
    – mp12345
    Jan 18, 2017 at 23:20
  • $\begingroup$ @mp12345 I updated, including avs's hint, in case you were wondering if it were necessary that $a_n$ and $b_n$ behave the way you predicted. $\endgroup$ Jan 18, 2017 at 23:22
  • $\begingroup$ thanks again for your help $\endgroup$
    – mp12345
    Jan 18, 2017 at 23:23
  • $\begingroup$ @mp12345 No problem :-) $\endgroup$ Jan 18, 2017 at 23:23

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