Does the series converge, problem with $(-1)^n$, but Leibnitz impossible. Hi my problem is connected with this series
$$
\sum_{n=2}^\infty \frac{(-1)^n}{\sqrt[3]n + (-1)^{n(n+1)/2}}
$$
I was trying to sum it by $4$ elements, but it didn't end up well.
I don't know how to deal with this problem, Leibnitz, Dirichlet, Abel is pointless here. Can you please help me?
 A: We consider the series
$$
\sum\limits_{n = 4}^{ + \infty } {\frac{{\left( { - 1} \right)^n }}
{{\sqrt[3]{n} + \left( { - 1} \right)^{\frac{{n\left( {n + 1} \right)}}
{2}} }}} 
$$
instead your series. If we prove that this series is convergent then your series is convergent also. We can rewrite the series as
$$
\sum\limits_{k = 1}^\infty  {\left( {a_{4k}  + a_{4k + 1}  + a_{4k + 2}  + a_{4k + 3} } \right)} 
$$
where
$$
a_{4k}  = \frac{1}
{{\sqrt[3]{{4k}} + 1}},a_{4k + 1}  =  - \frac{1}
{{\sqrt[3]{{4k + 1}} - 1}},a_{4k + 2}  = \frac{1}
{{\sqrt[3]{{4k + 2}} - 1}},a_{4k + 3}  =  - \frac{1}
{{\sqrt[3]{{4k + 3}} + 1}}
$$
Now, write down
$$
s_1 \left( k \right) = \frac{1}
{{\sqrt[3]{{4k}} + 1}} - \frac{1}
{{\sqrt[3]{{4k + 1}} - 1}}
$$
and rewrite is as function of $\frac{1}{k}$. After that using the Taylor series also, you can develop this function in powers of $\frac{1}{k}$. You will get
$$
s_1 \left( k \right) =  - \frac{{\left( {\frac{1}
{k}} \right)^{2/3} }}
{{\sqrt[3]{2}}} - \frac{{5\left( {\frac{1}
{k}} \right)^{4/3} }}
{{12\;2^{2/3} }} + \frac{{\left( {\frac{1}
{k}} \right)^{5/3} }}
{{12\sqrt[3]{2}}} + O\left( {\frac{1}
{{k^2 }}} \right)
$$
To be more clear, you have that
$$
\begin{gathered}
  s_1 \left( k \right) = \frac{1}
{{\sqrt[3]{{4k}} + 1}} - \frac{1}
{{\sqrt[3]{{4k + 1}} - 1}} \hfill \\
   \hfill \\
   = \frac{{\sqrt[3]{{4k + 1}} - \sqrt[3]{{4k}} - 2}}
{{\left[ {\sqrt[3]{{4k}} + 1} \right]\left[ {\sqrt[3]{{4k + 1}} - 1} \right]}} \hfill \\
   \hfill \\
   = \frac{{\sqrt[3]{{4k}}\left[ {\sqrt[3]{{1 + \frac{1}
{{4k}}}} - 1} \right] - 2}}
{{\left[ {\sqrt[3]{{4k}} + 1} \right]\left[ {\sqrt[3]{{4k}}\left( {\sqrt[3]{{1 + \frac{1}
{{4k}}}}} \right) - 1} \right]}} \hfill \\ 
\end{gathered} 
$$
and then, from here, you can use the Taylor expansion with
$$
{\sqrt[3]{{1 + \frac{1}
{{4k}}}}}
$$
Consider now
$$
s_2 \left( k \right) = \frac{1}
{{\sqrt[3]{{4k + 2}} - 1}} - \frac{1}
{{\sqrt[3]{{4k + 3}} + 1}}
$$
and repeat the above procedure. You will get
$$
s_2 \left( k \right) = \frac{{\left( {\frac{1}
{k}} \right)^{2/3} }}
{{\sqrt[3]{2}}} + \frac{{7\left( {\frac{1}
{k}} \right)^{4/3} }}
{{12\;2^{2/3} }} - \frac{{5\left( {\frac{1}
{k}} \right)^{5/3} }}
{{12\sqrt[3]{2}}} + O\left( {\frac{1}
{{k^2 }}} \right)
$$
Therefore
$$
s\left( k \right) = s_1 \left( k \right) + s_2 \left( k \right) = \frac{{\left( {\frac{1}
{k}} \right)^{4/3} }}
{{6 \cdot \;2^{2/3} }} - \frac{{\left( {\frac{1}
{k}} \right)^{5/3} }}
{{3\sqrt[3]{2}}} + O\left( {\frac{1}
{{k^2 }}} \right)
$$
so that
$$
s\left( k \right) = s_1 \left( k \right) + s_2 \left( k \right) = \frac{A}
{{k^{4/3} }} + O\left( {\frac{1}
{{k^{5/3} }}} \right)
$$
and the considered series is convergent as well as yours.
By the way: the above expansions which contain non integer powers of $\frac{1}{k}$ are called Puiseux expansions.
