I need to give an example of a two series sum $a_n$ and $b_n$ such that the $\lim_{n \to \infty}\frac{a_n}{b_n}=1$. One series has to diverge and one has to converge. $a_n$ and/or $b_n$ don't necessarily have to be positive. I have no idea where to start, because all of the tests we have learned so far, like ratio test and such, involve the series being both divergent or vice versa, so I'm stuck.

  • $\begingroup$ no i do mean two series $\endgroup$ Apr 9, 2014 at 2:36
  • $\begingroup$ So you want one of $\sum_n a_n$ and $\sum_n b_n$ to converge, the other to diverge, and require that the ratio of terms $a_n/b_n\rightarrow 1$, right? $\endgroup$
    – MPW
    Apr 9, 2014 at 2:38
  • $\begingroup$ yes that is correct, sorry if my question was worded incorrectly. $\endgroup$ Apr 9, 2014 at 2:39
  • $\begingroup$ Sorry for the incorrect answer (since deleted) didn't read you post closely enough. $\endgroup$ Apr 9, 2014 at 2:43

2 Answers 2



$$ a_n = \frac{(-1)^n}{\sqrt{n}} + \frac{1}{n} $$


$$ b_n = \frac{(-1)^n}{\sqrt{n} }$$


$$ \frac{a_n}{b_n} = 1 + \frac{\sqrt{n}}{n(-1)^n} = 1 + \frac{ (-1)^n}{\sqrt{n}} \to 1$$

But, you should check as an easy exercise that

$ \sum a_n $ diverges but $\sum b_n $ converges

  • $\begingroup$ i think this looks pretty good to me $\endgroup$ Apr 9, 2014 at 2:46
  • $\begingroup$ Why does $\sum a_n$ diverge? $\endgroup$ Apr 9, 2014 at 2:49
  • $\begingroup$ I know believe it (due to insufficient cancellation), but it seems to require an analysis of the (even) partial sums, which I would not call an easy exercise for a Calc II student to do on her/his own. $\endgroup$ Apr 9, 2014 at 2:55

$\require{cancel}$What about putting $\xcancel{a_n = 1/n^{1-\epsilon}}$ and $\xcancel{b_n = 1/n^{1+\epsilon}}$ for your favorite small positive $\epsilon$?

Edit: This answer is incorrect, sorry!

  • 1
    $\begingroup$ According to the limit comparison test, $a_n$ and $b_n$ cannot be always both positive!! $\endgroup$ Apr 9, 2014 at 2:45
  • $\begingroup$ Yeah, I blurted this out too quickly. It doesn't satisfy $a_n/b_n\rightarrow 1$, sorry :( $$ $$ I added strikeouts to my suggestion to indicate it doesn't work $\endgroup$
    – MPW
    Apr 9, 2014 at 2:46
  • $\begingroup$ I still appreciate the help $\endgroup$ Apr 9, 2014 at 17:57

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