Establish convergence of the series: $1-\frac{1}{2}-\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}-...$

We can see that the general term of this series is $$a_n=(-1)^n\sum_{k=\frac{n(n+1)}{2}+1}^{\frac{n(n+1)}{2}+n+1}\frac{1}{k}$$ and we have $$H_n=\sum_{k=1}^n\frac{1}{k}=\log n+\gamma+\frac{1}{2n}+O(\frac{1}{n^2})$$ so \begin{align}|a_n|=H_{\frac{n(n+1)}{2}+n+1}-H_{\frac{n(n+1)}{2}+1}&=\log(1+\frac{2}{n})+\frac{1}{(n+1)(n+2)}-\frac{1}{n(n+1)}+O(\frac{1}{n^2})\\ &=\frac{2}{n}+O(\frac{1}{n^2})\end{align} hence we have $$a_n=\frac{2(-1)^n}{n}+O(\frac{1}{n^2})$$ which allows us to conclude the convergence of the series since it's sum of two convergent series, one by alternating series test and the other by comparaison with Riemann series.

If I'm not getting it wrong, your series is

$$1-\frac 1 2 -\frac 13+\frac 14+\frac 15+\frac 16-\frac 17-\frac 18-\frac 19-\frac 1{10}+++++------\dots$$

So you have $1$ plus, $2$ minuses, $3$ pluses, $4$ minuses, and so on.

We can write your series as $a_0+a_1+a_2+a_3+\dots$ where $$\begin{align} a_0&=(-1)^0 1 \\ a_1&=(-1)^1\sum_{k=2}^{3}\frac 1 k\\ a_2&=(-1)^2\sum_{k=4}^{6}\frac 1 k\\ a_3&=(-1)^3\sum_{k=7}^{10}\frac 1 k\\ \cdots &=\cdots \\ a_n&=(-1)^n\sum_{k=T_n+1}^{T_{n+1}}\frac 1 k\end{align}$$

Where $T_n=\frac{n(n+1)}{2}$ so it goes $1,3,6,10,\dots$

Now, we know that $$\sum_{k=1}^n \frac 1 k =\log n+\gamma+\frac 1 {2n}+O(n^{-2})$$

Thus

$$\eqalign{ & \sum\limits_{k = 1}^{{T_n}} {{1 \over k}} = \log n + \log \left( {n + 1} \right) - \log 2 + \gamma + O({n^{ - 2}}) \cr & \sum\limits_{k = 1}^{{T_{n + 1}}} {{1 \over k}} = \log \left( {n + 2} \right) + \log \left( {n + 1} \right) - \log 2 + \gamma + O({n^{ - 2}}) \cr} $$

Whence, after simplification

$$\sum\limits_{k = {T_n} + 1}^{{T_{n + 1}}} {{1 \over k}} = \log \left( {1 + {2 \over n}} \right) + O(n^{-2})$$

Recall that $$\log(1+x)=x+O(x^2)$$ so

$$\sum\limits_{k = {T_n} + 1}^{{T_{n + 1}}} {{1 \over k}} = \frac{2}{n} +O\left(\frac 1 {n^2}\right)$$

Since $$\sum (-1)^n \frac 1 n $$ and $$\sum n^{-2}$$ converge, so does your series.

You can use the alternating series test if you can prove that the sum of a block goes to zero. Block $n$ starts at $\dfrac 1{\dfrac {n(n-1)}2+1}$ and ends at $\dfrac 1{\frac {n(n+1)}2}$ and has $n$ terms. The sum is then less than $\dfrac n{\dfrac {n(n-1)}2}=\dfrac 2{n-1}$ which goes to zero

$$ \underbrace{\vphantom{\frac11}+1}_{\text{length }1} \underbrace{-\frac12-\frac13}_{\text{length }2} \underbrace{+\frac14+\frac15+\frac16}_{\text{length }3} \underbrace{-\frac17-\frac18-\frac19-\frac1{10}}_{\text{length }4} \underbrace{+\frac1{11}+\frac1{12}+\frac1{13}+\frac1{14}+\frac1{15}}_{\text{length }5}-\ldots $$ The absolute values of the terms of the same-sign block of length $n$ are from $$ \dfrac1{n(n-1)/2+1}\quad\text{to}\quad\dfrac1{n(n+1)/2} $$ and the sum of the block must satisfy $$ \frac2{n+1}=\frac{n}{n(n+1)/2}\le(-1)^{n-1}\text{sum}\le\frac{n}{n(n-1)/2+1}\lt\frac2{n-1} $$ which tends to $0$.

The absolute value of the sum of the same-sign block of length $n$ and the same-sign block of length $n+1$ is at most $$ \frac2{n-1}-\frac2{n+2}=\frac6{(n+2)(n-1)} $$ Thus, the sum of pairs of blocks converge absolutely and the blocks converge to $0$. Thus, the full series converges.