Prove convergence of series Let $$\displaystyle a_n=\sum_{n=2}^{\infty} \frac{(-1)^n}{n^{\frac{1}{3}}+(-1)^{\frac{n(n+1)}{2}}}$$ 
so I divide it into four series $4k, 4k+1, 4k+2, 4k+3$ and I pair for instance $4k, 4k+1$ and $4k+2, 4k+3$ and prove that these two series is convergent and conclude that since 
both series are convergent so the sum of it is also convergent but I'm not sure if it's legall 
second series I'm not sure is like this $\displaystyle b_n=\frac{1}{1}+\frac{1}{3}-\frac{1}{2}+\frac{1}{5}+\frac{1}{7}-\frac{1}{4}+\frac{1}{9}+\frac{1}{11}-\frac{1}{6}+...$ 
here I also split it into smaller parts namely $\displaystyle \sum_{n=1}^{\infty}\frac{1}{4n-3}+\sum_{n=1}^{\infty}\frac{1}{4n-1}-\sum_{n=1}^{\infty}\frac{1}{2n}$ and I prove that $\displaystyle \frac{1}{4n-3}+\frac{1}{4n-1}-\frac{1}{2n}<\frac{1}{n^2} \cdot c $ for some const c so it implies that origin series is convergent or not ? 
 A: Hint. Here is an approach. 
Recall that, for $x$ near $0$, you have by the Taylor series expansion or simply by the finite sum of a geometric sequence:
$$
\frac{1}{1+x}=1-x+x^2+\mathcal{O}(x^3),
$$
You may then write, as $n$ is great:
$$
\frac{(-1)^n}{n^{1/3}+(-1)^{\frac{n(n+1)}{2}}}=\frac{(-1)^n}{n^{1/3}\left(1+\frac{(-1)^{\frac{n(n+1)}{2}}}{n^{1/3}}\right)}=\frac{(-1)^n}{n^{1/3}}-\frac{(-1)^{\frac{n(n+3)}{2}}}{n^{2/3}}+\mathcal{O}\left(\frac{1}{n^{4/3}}\right)
$$ The series $ \displaystyle
\sum\frac{(-1)^n}{n^{1/3}}$
 is convergent by the alternating series test,  the series $\displaystyle
\sum\frac{1}{n^{4/3}}$ is convergent $(4/3>1)$. Since
$$
\left|\sum_{k=2}^{2N}(-1)^{n(n+3)/2}\right|\leq 2, \qquad \left|\sum_{k=2}^{2N+1}(-1)^{n(n+3)/2}\right|\leq 2,\quad N=1,2,3\ldots,
$$ then the series $ \displaystyle
\sum\frac{(-1)^{\frac{n(n+3)}{2}}}{n^{2/3}}$
 is convergent by the Dirichlet series test.
A: Consider the series $\displaystyle\sum_{m=1}^{\infty} a_m=\sum_{n\in A}(-1)^n\frac{1}{n^{1/3}-1}=\frac{1}{2^{1/3}-1}-\frac{1}{5^{1/3}-1}+\frac{1}{6^{1/3}-1}-\frac{1}{9^{1/3}-1}+\cdots$ $\hspace{.8 in}$and $\displaystyle\sum_{m=1}^{\infty}b_m=\sum_{n\in B}(-1)^n\frac{1}{n^{1/3}+1}=-\frac{1}{3^{1/3}+1}+\frac{1}{4^{1/3}+1}-\frac{1}{7^{1/3}+1}+\frac{1}{8^{1/3}+1}-\cdots$,  
$\hspace{.4 in}$where $A=\{n\in\mathbb{N}, n\equiv1 \text{ or } n\equiv2 \pmod 4 \ \text { and } n> 1\}$ and
$\hspace{.8 in}B=\{n\in\mathbb{N}, n\equiv3 \text{ or } n\equiv4 \pmod 4 \}$.
Then both series converge by the Alternating Series Test; so if we let 
$(T_m), (U_m), \text{ and }(S_m)$ be the partial sums for $\displaystyle\sum_{m=1}^{\infty} a_m, \displaystyle\sum_{m=1}^{\infty} b_m$, and the given series, respectively,
then $T_m\to T$ and $U_m\to U$ for some numbers $T$ and $U$.
Since  $S_{2m}=T_m+U_m$ for every $m$, $\;\;S_{2m}\to T+U$; 
and since $S_{2m+1}=S_{2m}+\frac{1}{(2m+2)^{1/3}\pm1}, \;\;S_{2m+1}\to T+U$.
Therefore $S_m\to T+U$, and the series converges.
