Determine if the series converges absolutely, converges conditionally, or diverges. Determine if the series converges absolutely, converges conditionally, or diverges. Find the exact value for the sum of the convergent series. $$ 1-\frac{1}{2} - \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} - \frac{1}{8} - \frac{1}{9} ... - \frac{1}{15} + \frac{1}{16} + ... + \frac{1}{31} - ... $$
That is, one positive term, followed by two negative, four positive, eight negative, etc.
I am utterly and completely lost here. My instinct is to compare it to another series, but I am not sure which to compare it to. I don't think I can use the alternate harmonic series since it does not alternate between positive and negative for every other term. It "alternates" but in a different way here.
Thank you so much for your help in advance.
 A: Let $H_n = 1 + \frac{1}{2} + \ldots + \frac{1}{n}$. Then from standard estimates we have that
$$ \ln(n+1) = \int_{1}^{n+1} \frac{1}{x}dx \le H_n \le 1 + \int_{1}^n \frac{1}{x} dx = 1 + \ln(n).$$
In fact as shown in https://www.math.drexel.edu/~tolya/123_harmonic.pdf, we have that
$$\lim_{n \to \infty} \underbrace{H_n - \ln(n)}_{\delta_n} \to \gamma \approx 0.5772.$$
Now let us look at one of the blocks of the sequences from $(2^k - 1)$st term to the $(2^{k+1}-1)$st term. We know that all of the terms in one such block have the same sign. Let $a_k$ be the sum of the elements in the $k$ block. Then we have that
$$|a_k| = H_{2^{k+1}-1}-H_{2^k-1} = \delta_{2^{k+1}-1}-\delta_{2^k-1} + \ln(2^{k+1}-1) - \ln(2^k-1).$$
Now if, we send $k$ to infinity, we see that $|a_k| \to \ln(2)$. Thus, the alternating series $a_k$ diverges.
Now let $S_n$ be the partial series of the series in the question. The series in the question converging means that $S_n$ converges as a sequence. However, in that case every subsequence of $S_n$ converges. However, we just found a subsequence that diverged. Hence our original series diverges.
