Variation on Harmonic Series

Question

I'm trying to establish the convergence or divergence of the following variant of the harmonic series: $$\frac{1}{1}+\frac{1}{2}-\frac{1}{3}-\frac{1}{4}-\frac{1}{5}+\frac{1}{6}+\frac{1}{7}-\frac{1}{8}-\frac{1}{9}-\frac{1}{10}+\frac{1}{11}\cdots$$

Where the sign pattern has period 5, ie, it looks like this: ++---++---++---....

My thought has been to find a regrouping that diverges, since I only need to find one in order to show the series is divergent. I tried to bound this series below by increasing the denominator on the positive terms, and decreasing it on the negative terms to yield a series like $$\left(\frac{1}{2}+\frac{1}{2}\right)-\left(\frac{1}{3}+\frac{1}{3}+\frac{1}{3}\right)+\left(\frac{1}{7}+\frac{1}{7}\right)+ \cdots$$

but I'm pretty sure this will converge. I don't really know what approach to take next. A hint would be greatly appreciated! Thanks!

Answer 1

The sum of the first two is $\dfrac1{5n+1}+\dfrac1{5n+2} \lt \dfrac{2}{5n} $ and the sum of the last 3 is $\dfrac1{5n+3}+\dfrac1{5n+4}+\dfrac1{5n+5} \gt \dfrac{3}{5n+5} $.

Therefore

$\begin{array}\\ \dfrac1{5n+1}+\dfrac1{5n+2} -(\dfrac1{5n+3}+\dfrac1{5n+4}+\dfrac1{5n+5}) &\lt \dfrac{2}{5n}-\dfrac{3}{5n+5}\\ &=\dfrac{2 - n}{5 n (n + 1)}\\ &=\dfrac{2}{5 n (n + 1)}-\dfrac{1}{5 (n + 1)}\\ \end{array} $

and the sum of this diverges to $-\infty$.

Answer 2

Denote your sequence as $\sum a_k $. Consider the regrouping $\sum b_k$ where $$ b_k = a_{5k-4} + a_{5k-3} + a_{5k-2} + a_{5k-1} + a_{5k} $$ $$ = \frac{1}{5k-4} + \frac{1}{5k-3} - \frac{1}{5k-2} - \frac{1}{5k-1} -\frac{1}{5k} $$ Combine the terms using the common denominator $(5k-4)(5k-3)(5k-2)(5k-1)(5k)$. This yields a polynomial of degree four divided by a polynomial of degree five. Polynomial long division gives a polynomial of degree zero over a polynomial of degree one. Use the limit comparison test with the harmonic series $\sum_{n=1}^{\infty} \frac{1}{n}$ to conclude that the regrouping $\sum b_k$ is divergent.

Every regrouping of a convergent sequence is itself convergent. Since $\sum b_k$ diverges, we can conclude $\sum a_k$ diverges as well.