Series converges or diverges?

Question

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 .

Answer 1

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.

Answer 2

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.

Answer 3

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.