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 ? 

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.
