Consider the series $$\sum_{n=1}^{\infty}\frac{(-1)^{\frac{n(n+1)}{2}+1}}{n}=1+\dfrac12-\dfrac13-\dfrac14+\dfrac15+\dfrac16-\cdots.$$ This is clearly not absolutely convergent. On the other hand, obvious choice, alternating series does not work here. Seems like the partial sum sequence is bounded but it is not monotone.
How can we prove that this series converges? and, where does it converge to?
On
The triangular numbers alternate odd, odd, even, even, odd, odd, even, even, etc. The reason is that to move from $T_n$ to $T_{n+2}$, we add $n + (n + 1) = 2n + 1$, an odd number, so we obtain an alternating pattern of parities for $T_n$, over the odd and even integers $n$.
So, consider grouping the terms in pairs: $$\left(1+\frac12\right)-\left(\frac13+\frac14\right)+\left(\frac15+\frac16\right)-\ldots = \sum_{n=1}^\infty (-1)^n\left(\frac{1}{2n-1} + \frac{1}{2n}\right).$$ This series is convergent, using the alternating series test.
Because the terms of the original series converge to $0$, this also implies that the original series is convergent too.
On
The triangle numbers themselves follow the pattern of
$$odd, odd, even, even, odd, odd, even, even, \ldots$$
This can be shown as this pattern alternates whether or not the even number in $\frac{n(n+1)}{2}$ is $0,2 \mod 4$. So as you can see in your sum, the terms will alternate in a pattern of twos.
Now if we group these terms of the series as follows
$$ (1+1/2)-(1/3+1/4) +(1/5+1/6) - \ldots = \sum_{n=1}^\infty \left(\dfrac{1}{2n-1}-\dfrac{1}{2n}\right) $$
which converges via the alternating series test, and according to Wolfram alpha, it converges to the value of $$ \sum_{n=1}^\infty \left(\dfrac{1}{2n-1}-\dfrac{1}{2n}\right) = \dfrac{1}{4}(\pi+2\log(2)) $$
However you can only associate the series if it converges, so that associative step we made assumes that your initial series converges. I'd expect your initial series to converge, however I am much to tired and need to sleep!
On
The sequence of partial sums is $s_n=s_{n-1}+\frac{(-1)^{T_n+1}}{n}$, and if we can partition the sequence into two subsequences which converge to the same value, it implies the original sequence converges as well. This is usually done by looking at subsequences made by odd and even indeces, see for example If $\{x_{2m}\}$ and $\{x_{2m-1}\}$ converge to the same limit, does $\{x_m\}$ converge?. Notice that we don't need to care about terms re-arrangements since we will be working only with finite partial sums.
So we consider two subsequences by splitting the $s_n$ for and even $s_{2n}$ and odd $s_{2n-1}$ subsequences. Since trinagular numbers parity is periodic modulo $4$, we can write these explicitly in terms of $s_{4n},s_{4n-2}$ and $s_{4n-1},s_{4n-3}$. For example for the even indexes subsequence, say $b_{n}$ such that $b_{2n}=s_{4n}$ and $b_{2n-1}=s_{4n-2}$, applying the recurrence above we can see that $$ b_{2n}=b_{2n-1}-\frac{1}{4n-1}-\frac{1}{4n}, b_{2n-1}=b_{2n-2}+\frac{1}{4n-3}+\frac{1}{4n-2}\\ $$ which together with $b_1=s_2=\frac{3}{2}$ describes the sequence completely. But this is indeed the sum $$b_n=\sum_{k=1}^{\infty}(-1)^{k+1} \left(\frac{1}{2k-1}+\frac{1}{2k}\right).$$ Similarly for the sequence $a_n$ of odd terms $a_{2n}=s_{4n-1}$, $a_{2n-1}=s_{4n-3}$, we arrive at
$$ a_{2n}=a_{2n-1}+\frac{1}{4n-2}-\frac{1}{4n-1},a_{2n-1}=a_{2n-2}+\frac{1}{4n-4}-\frac{1}{4n-3} $$ with $a_1=s_1=1$, corresponding to $$a_n=1+\sum_{k=2}^{\infty}(-1)^{k} \left(\frac{1}{2k-2}+\frac{1}{2k-1}\right).$$ Now both sequences clearly converge by the alternating series test, so we are left with showing they converge to the same value. But since both of the sums are themselves just added $\frac{\pi}{4}=\sum_{k=0}^{\infty} \frac{(-1)^k}{2k+1}$ (Leibniz formula) and taylor series for natural logarithm $\ln(1+x)=\sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{k}x^k$ for $x=1$. So, as already shown in the other answer, indeed $$ \lim_{n \to \infty} s_n = \lim_{n \to \infty} a_n = \lim_{n \to \infty} b_n = \frac{\pi}{4}+\frac{\ln 2}{2}. $$
On
$$1\color{red}{+\frac12}\color{blue}{-\frac13}-\frac14+\frac15\color{cyan}{+\frac16}\color{magenta}{-\frac17}+\cdots$$
$$=1\color{red}{-\frac12+2\cdot\frac12}\color{blue}{+\frac13-2\cdot\frac13}-\frac14+\frac15\color{cyan}{-\frac16+2\cdot\frac16}\color{magenta}{+\frac17-2\cdot\frac17}+\cdots$$
$$=1-\frac12+\frac13-\frac14+\frac15-\cdots+2\left(\frac12-\frac13+\frac16-\frac17+\cdots\right)$$
$$=\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n}+2\sum_{n=1}^\infty\frac{1}{4n-2}-\frac{1}{4n-1}$$
$$=\ln(2)+\frac12\sum_{n=1}^\infty\frac{1}{n-1/2}-\frac{1}{n-1/4}$$
$$=\ln(2)+\frac12\left(H_{-1/4}-H_{-1/2}\right)$$
$$=\ln(2)+\frac12\left(\frac{\pi}{2}-\ln(2)\right)$$
$$=\frac{\pi}{4}+\frac12\ln(2)$$
note that we used the series representation of the harmonic number
$$H_a=\sum_{n=1}^\infty\frac{1}{n}-\frac{1}{n+a}\Longrightarrow H_a-H_b=\sum_{n=1}^\infty\frac{1}{n+b}-\frac{1}{n+a}$$ and we also used the results $H_{-1/4}=\frac{\pi}{2}-3\ln(2)$ and $H_{-1/2}=-2\ln(2)$ which can be obtained from the integral representation of the harmonic number $H_a=\int_0^1\frac{1-x^a}{1-x}dx$.
Observe that $$ \left| {\sum\limits_{n = 1}^N {( - 1)^{T_n + 1} } } \right| \le 2 $$ for any $N\geq 1$. Also $\frac{1}{n}\to 0$ monotonically. Hence, by Dirichlet's test, the series converges. The limit is \begin{align*} \sum\limits_{n = 1}^\infty {\frac{{( - 1)^{T_n + 1} }}{n}} & = \mathop {\lim }\limits_{N \to + \infty } \sum\limits_{n = 1}^{2N} {\frac{{( - 1)^{T_n + 1} }}{n}} = \mathop {\lim }\limits_{N \to + \infty } \left( {\sum\limits_{n = 1}^N {\frac{{( - 1)^{n + 1} }}{{2n - 1}}} + \sum\limits_{n = 1}^N {\frac{{( - 1)^{n + 1} }}{{2n}}} } \right) \\ & = \mathop {\lim }\limits_{N \to + \infty } \sum\limits_{n = 1}^N {\frac{{( - 1)^{n + 1} }}{{2n - 1}}} + \mathop {\lim }\limits_{N \to + \infty } \sum\limits_{n = 1}^N {\frac{{( - 1)^{n + 1} }}{{2n}}} = \frac{\pi }{4} + \frac{1}{2}\log 2. \end{align*}