Does the series converge?

$$ 1+\frac 1 {3^2}-\frac12 +\frac1 {5^2} + \frac 1{7^2} - \frac14 \cdots + \frac1{(4n+1)^2} + \frac1{(4n+3)^2} - \frac 1{2n+2}+\cdots$$

I can see that the series does not converge absolutely by comparison of $\sum1/(2n+2)$ with $\sum 1/n$ but this does not imply anything about its conditional convergence.

Also, if it converges conditionally, then its rearrangements will converge to any real number including $\infty$ ($\pm\infty$).

Please suggest a test using comparison/root test/ratio test since I am following Rudin alone .


It cannot converge, since $\sum (\frac{1}{(4n+1)^2}+\frac{1}{(4n+3)^2})$ converges but $\sum \frac{1}{2n+2}$ diverges, the sum of the two series is doomed to diverge.

  • $\begingroup$ Yeah but that is true for absolute convergence. What about conditional convergence? $\endgroup$ – Silver moon Nov 5 '14 at 21:19
  • 2
    $\begingroup$ It's true for conditional convergence. If $\sum a_n$ diverges and $\sum b_n$ converges, then $\sum a_n+b_n$ diverges (conditionally). $\endgroup$ – Pierre Alvarez Nov 5 '14 at 21:21
  • $\begingroup$ It just doesn't matter, what I say is still true. The sum of two divergent series can converge, but divergent+convergent=divergent. $\endgroup$ – Pierre Alvarez Nov 5 '14 at 21:24
  • $\begingroup$ What you have written is a rearrangement of the original series. It cannot be guaranteed that if a rearrangement diverges, other rearrangements also diverge. What we can be sure about is that if a series converges absolutely then all its rearrangements converge to the same value, but it is not the case with this series, since the series diverges absolutely. $\endgroup$ – Silver moon Nov 5 '14 at 21:30
  • $\begingroup$ No it's not a rearrangement. I only wrote it as the sum of two series. Just consider the partial sums as two sequences of which the limit of the sum is your serie. Well it's clear that if $(a_n)_n$ diverges and $(b_n)_n$ converges then $(a_n+b_n)_n$ diverges. $\endgroup$ – Pierre Alvarez Nov 5 '14 at 21:32

For conditional convergence (but not absolute convergence) you must have divergence of both the series of positive terms and the series of negative terms. As Pierre pointed out: in this case you have convergence of one but not of the other.

  • $\begingroup$ Thank you. I hope that some support might be able to make Martin trust what I say. $\endgroup$ – Pierre Alvarez Nov 5 '14 at 21:41
  • $\begingroup$ @Pierre: I assume you are correct, but I am very confused at this point with rearrangements $\endgroup$ – Silver moon Nov 5 '14 at 21:50
  • $\begingroup$ @GEdgar: Did you mean convergence in your comment? "For conditional convergence (but not absolute convergence) you must have divergence of both" $\endgroup$ – Silver moon Nov 5 '14 at 21:52

In $\frac1{(4n+1)^2} + \frac1{(4n+3)^2} - \frac 1{2n+2}$, the last term swamps the other terms, since its sum diverges while the sum of the others converges.

However, if the terms are $\frac1{(4n+1)^2} + \frac1{(4n+3)^2} - \frac 2{(4n+2)^2}$, things get trickier.

$\begin{array}\\ d(n) &=\frac1{(4n+1)^2} + \frac1{(4n+3)^2} - \frac 2{(4n+2)^2}\\ &=\frac{(4n+3)^2(4n+2)^2+(4n+2)^2(4n+1)^2-2(4n+3)^2(4n+1)^2}{(4n+1)^2(4n+2)^2(4n+3)^2}\\ &=\frac{(4n+3)^2(4n+2)^2-(4n+3)^2(4n+1)^2+(4n+2)^2(4n+1)^2-(4n+3)^2(4n+1)^2}{(4n+1)^2(4n+2)^2(4n+3)^2}\\ &=\frac{((4n+3)(4n+2)-(4n+3)(4n+1))((4n+3)(4n+2)^2+(4n+3)(4n+1))+((4n+2)(4n+1)-(4n+3)(4n+1))((4n+2)(4n+1)-(4n+3)(4n+1))}{(4n+1)^2(4n+2)^2(4n+3)^2}\\ &=\frac{((16n^2+20n+6)-(16n^2+16n+3))((16n^2+20n+6)+(16n^2+16n+3))+((16n^2+12n+2)-(16n^2+16n+3))(16n^2+12n+2)+(16n^2+16n+3))}{(4n+1)^2(4n+2)^2(4n+3)^2}\\ &=\frac{(4n+3)(32n^2+36n+9)+((-4n-1)(32n^2+28n+5))}{(4n+1)^2(4n+2)^2(4n+3)^2}\\ \end{array} $

Looking at the numerator,

$\begin{array}\\ (4n+3)(32n^2+36n+9)-((4n+1)(32n^2+28n+5)) &=(128n^3+(32\cdot 3 + 4\cdot 36)n^2+O(n)) -(128n^3+(32+4\cdot 28)n^2+O(n))\\ &=(96 + 144-32-112)n^2+O(n))\\ &=96n^2+O(n)\\ \end{array} $

so $d(n) = \frac{96n^2+O(n)}{(4n+1)^2(4n+2)^2(4n+3)^2} $

and the sum of these converges.

P.S. If I haven't made any algebra errors in this, I will be surprised, but the final conclusion about the convergence of the modified sum should be correct.


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