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.
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.