Convergence of $\sum \frac{b_n}{n}$ Where $b_n = 1, -1, -1, 1, 1, 1, -1, -1,-1,-1,1,1,1,1,1,....$ How can I show convergence or divergence of the following sequence?
$$\sum \frac{b_n}{n}$$
Where $b_n = 1, -1, -1, 1, 1, 1, -1, -1,-1,-1,1,1,1,1,1,....$
Im not sure how to use any of the standart theorems to show convergence or divergence.
Any attempts to use comparison test, or finding an upper bound failed.
Any ideas?
 A: This is equivalent to whether: $$\sum_{n=0}^\infty (-1)^n c_n$$ converges, where $$c_n=\sum_{k=T_{n}+1}^{T_{n+1}}\frac1{k}=\sum_{j=1}^{n+1}\frac{1}{T_n+j}$$
Where $T_n=1+2+\cdots +n=\frac{n(n+1)}{2}.$
Then $$\begin{align} c_n-c_{n+1}&=\sum_{j=1}^{n+1}\left(\frac{1}{T_n+j}-\frac1{T_{n+1}+j}\right)-\frac1{T_{n+2}}\\&=\sum_{j=1}^{n+1}\frac{n+1}{(T_{n}+j)(T_{n}+n+1+j)}-\frac{2}{(n+2)(n+3)}\end{align} 
$$
Now, $$\frac{n+1}{(T_n+j)(T_n+n+1+j)}\geq \frac{n+1}{T_{n+1}T_{n+2}}=\frac{4}{(n+2)^2(n+3)}$$ for $j\leq n+1,$ so
$$c_{n}-c_{n+1} \geq \frac{4(n+1)}{(n+2)^2(n+3)}-\frac2{(n+2)(n+3)}=\frac{2n}{(n+2)^2(n+3)}\geq 0$$
So $c_n$ is a decreasing positive sequence, and $\sum (-1)^nc_n$ converges. (Technically, you need to prove $c_n\to0,$ but this is relatively easy.)
If $S_N=\sum_{n=1}^{N}\frac{b_n}n$, we have shown $S_{T_k}$ converges as $k\to\infty.$
But this means your series converges because for each $n,$ there is a $k$ such that $T_k\leq n< T_{k+1},$ and then $S_n$ is between $S_{T_k}$ and $S_{T_{k+1}},$ and as $n\to\infty,$ $k\to\infty.$

Addendum
More generally:

Lemma: If $a_n$ is a non-zero sequence, and $K_1,K_2,\dots $ is an increasing sequence of natural numbers with $a_{K_i+1},\dots,a_{K_{i+1}}$ of the same sign, then $S_{N}=\sum_{n=1}^{N}a_n$ converges as $N\to\infty$ if and only if $S_{K_i}$ converges as $i\to\infty.$

(We definitely don’t need $a_n$ non-zero here, just the language becomes more complicated to allow zero values.)
Note, we don’t need that the sign of $a_n$ switches between $a_{K_{i+1}}$ and $a_{K_{i+1}+1}.$ That is the most useful and interesting case, but what this Lemma is saying is we can combine any finite sets of contiguous terms of the same sign in our sum and have the same convergence question.
The essential proof of the lemma is that if $K_i<n\leq K_{i+1}$ then: $$\min(S_{K_i},S_{K_{i+1}})\leq S_n\leq \max(S_{K_i},S_{K_{i+1}})$$ and the limit of the minimum and maximum converge to the same value.
A: The $n$-th "run" consists of $n$ elements (numerators of $1$ for odd $n$ and numerators of $-1$ for even $n$). The last element of the $n$-th run has index $n(n+1)/2$, and the first element thus has index $n(n+1)/2 - n+1$
The respective run adds
$$
r_n = \!\!\!\!\!\!\sum_{k=n(n+1)/2-n+1}^{n(n+1)/2} \!\!\!\!\!\frac {(-1)^{n+1}}k = (-1)^{n+1} (H_{n(n+1)/2} - H_{n(n-1)/2})
\tag 1$$
to the value $S$ of the series, where $H_n$ is the $n$-th Harmonic number with $H_0:=0$ and
$$H_n\approx \ln n +\gamma+\frac1{2n}\tag 2$$
The complete sum is then
$$\begin{align}
S &= r_1 + \sum_{n=2}^\infty r_n \\
&\stackrel{(1)}= H_1 + \sum_{n=2}^\infty (-1)^{n+1} (H_{n(n+1)/2} - H_{n(n-1)/2}) \\
&\stackrel{(2)}\approx 1 - \sum_{n=2}^\infty (-1)^{n}\left(\ln(n(n+1)/2)-\ln(n(n-1)/2)+\frac1{n(n+1)}-\frac1{n(n-1)} \right) \\
&= 1 - \sum_{n=2}^\infty (-1)^{n}\left(\ln\frac{n+1}{n-1}  -\frac{2}{n^2-1} \right) \\
&= 1 - \sum_{n=2}^\infty (-1)^{n}\Bigg(\underbrace{\ln\left(1+\frac{2}{n-1}\right)}_{\textstyle\approx 2/(n-1)}  -\frac{2}{n^2-1} \Bigg) \\
\end{align}$$
This is to be understood as a sketch that has to be fleshed out and be made rigorous, but from the last line it should be clear that the series converges.
A: Let $S_k=b_1+b_2+\ldots +b_k.$
Denote $$u_n={2n(2n-1)\over 2}=n(2n-1)\quad  v_n={(2n+1)2n\over 2}=(2n+1)n$$ Then
$$S_{u_n}=n,\qquad S_{v_n}=-1$$
Moreover $S_k$ is decreasing for $u_n\le k\le v_n$ and increasing for $v_n\le k\le u_{n+1}.$
Therefore
$$|S_k|\le n+1,\qquad u_n\le k\le u_{n+1}$$
The condition $k\ge u_n$ implies $n\le \sqrt{2k}$ hence$$|S_k|\le \sqrt{2k}+1$$
We obtain (setting $S_0=0$)
$$\sum_{k=1}^N{b_k\over k}=\sum_{k=1}^N{S_k-S_{k-1}\over k}= \sum_{k=1}^{N-1}{S_k\over k(k+1)}+{S_N\over N} $$
The right hand side is convergent when $N\to \infty.$ So the series is convergent. Moreover we can estimate the rate of convergence
$$\left |\sum_{n=N}^\infty {b_n\over n}\right |\le {12\over \sqrt{N}}$$
A: See OEIS A353874 for the decimal expansion of your sum which indicates

There are n terms in the n-th group $v(n)$, from $1 / ((n^2-n+2)/2)$ up to $1 / ((n^2+n)/2)$.

and

As $|v(n+1)| < |v(n)|$, this series is convergent according to the alternating series test.

Your sum can also be represented as
$$\sum\limits_{n=1}^\infty (-1)^{n-1} \sum\limits_{k=0}^{n-1} \frac{1}{\frac{n\, (n-1)}{2}+1+k}=\sum\limits_{n=1}^\infty (-1)^{n-1} \left(H_{\frac{n\, (n+1)}{2}}-H_{\frac{n\, (n-1)}{2}}\right)$$
so $v(n)$ can be defined as
$$v(n)=(-1)^{n-1} \sum\limits_{k=0}^{n-1} \frac{1}{\frac{n\, (n-1)}{2}+1+k}=(-1)^{n-1} \left(H_{\frac{n\, (n+1)}{2}}-H_{\frac{n\, (n-1)}{2}}\right)$$
