Convergence of the series $\sum_n (-1)^{\lfloor \sqrt{n-1}\rfloor} \frac 1 n$ Let's consider the series $\sum a_n$, with $$a_n = (-1)^{\lfloor \sqrt{n-1}\rfloor} \frac 1 n$$
It looks absolutely like an example for the Leibniz test but here the signs don't interchange one for one. How can we prove the convergence of this series?

You can reduce the question to a question about Liebniz series by inserting parenthesis around the terms that have the same sign. The original series $\sum_{n=1}^{\infty} a_n$ is:
$$ \frac{1}{1} - \frac{1}{2} - \frac{1}{3} - \frac{1}{4} + \frac{1}{5} + \dots + \frac{1}{9} - \frac{1}{10} - \dots. $$ 
Denote the partial sums of the series $\sum_{n=1}^{\infty} a_n$ by $S_n = \sum_{k=1}^n a_k$. Inserting parenthesis around the terms with the same sign, we obtain a new series $\sum_{n=1}^{\infty} b_n$:
$$ \frac{1}{1} - \left( \frac{1}{2} + \frac{1}{3} + \frac{1}{4} \right) + \left( \frac{1}{5} + \dots + \frac{1}{9} \right) - \dots $$
Formally, what we do is we define $b_n$ by letting
$$ c_n = \sum_{k=(n-1)^2 + 1}^{n^2} \frac{1}{k}, \\
   b_n = (-1)^{n+1} c_n = (-1)^{n+1} \sum_{k=(n-1)^2 + 1}^{n^2} \frac{1}{k}. $$
Denote by $T_n$ the partial sums of the series $\sum_{n=1}^{\infty} b_n$. The fact that the new series $\sum_{n=1}^{\infty} b_n$ is obtained by inserting parenthesis around the terms corresponds to the fact that the sequence of partial sums of $\sum_{n=1}^{\infty} b_n$ is a subsequence of the sequence of partial sums of $\sum_{n=1}^{\infty} a_n$. Namely, $T_n = S_{n^2}$. If we can show that $c_n$ are monotonically decreasing to zero, we will obtain that $\sum_{n=1}^{\infty} b_n$ is a Liebniz series and hence converges. This shows that $S_{n^2}$ has a limit. Since the series $\sum_{n=1}^{\infty} b_n$ was obtained from the series $\sum_{n=1}^{\infty} a_n$ by inserting parenthesis around terms that have the same sign, this implies that $S_n$ also has a limit (the same limit).
Estimating $c_n$, we have
$$ c_n = \sum_{k=(n-1)^2 + 1}^{n^2} \frac{1}{k} \leq \sum_{k=(n-1)^2 + 1}^{n^2} \frac{1}{(n-1)^2 + 1} = \frac{n^2 - (n-1)^2}{(n-1)^2 + 1} = \frac{2n - 1}{(n-1)^2 + 1} \xrightarrow[n \to \infty]{} 0 $$
so $c_n$ tends to zero.
Finally,
$$ c_n - c_{n+1} = \sum_{k=(n-1)^2 + 1}^{n^2} \frac{1}{k} - \sum_{k=n^2 + 1}^{(n+1)^2} \frac{1}{k} \\ 
= \sum_{i = 0}^{2n - 2} \left(  \frac{1}{(n-1)^2 + 1 + i} - \frac{1}{n^2 + 1 + i} \right) - \frac{1}{(n+1)^2 - 1} - \frac{1}{(n+1)^2} \\
= \sum_{i = 0}^{2n - 2} \frac{2n - 1}{\left( (n-1)^2 + 1 + i \right) \left( n^2 + 1 + i \right)} - \frac{1}{(n+1)^2 - 1} - \frac{1}{(n+1)^2} \\
\geq  \frac{(2n-1)^2}{\left( (n-1)^2 + 1 + (2n-2) \right) \left( n^2 + 1 + (2n-2) \right)} - \frac{1}{(n+1)^2 - 1} - \frac{1}{(n+1)^2} \\
\geq \frac{(2n-1)^2}{n^2(n^2 + 2n - 1)} - \frac{2}{(n^2 + 2n - 1)^2} \\
= \frac{2(n-1)^2 - 1}{(n^2 + 2n + 1)^2} > 0 $$
for $n > 1$ which shows that $c_n$ are monotonically decreasing.
