$\sum a_k$ converges iff $\sum a_{2k}+a_{2k+1}$ converges A more clear statement:
Suppose $a_k \to 0$.  The series $\sum_{k=1}^{\infty} a_k$ converges iff $\sum_{k=1}^{\infty} (a_{2k}+a_{2k+1})$ converges.
I have been able to prove the forward direction.  I'm stuck on backward direction.
 A: Hint: Because $\sum (a_{2k}+a_{2k+1})$ converges, the sequence of partial sums of this series is Cauchy.
Conclude from this that the sequence of partial sums of the series $\sum a_n$ is Cauchy.  For odd $m$ and/or even $n$, where $m\lt n$,  we will need the fact that for large $m$ and $n$, the terms $a_m$ and $a_n$ have absolute value close to $0$. This part follows from the assumption that the $a_n$ have limit $0$. 
A: Suppose that $\sum_{k\ge 1}(a_{2k}+a_{2k+1})=L$; clearly we want to show that $\sum_{k\ge 1}a_k=L+a_1$. Let $\epsilon>0$; there is an $m_\epsilon\in\Bbb N$ such that $$\left|L-\sum_{k=1}^n(a_{2k}+a_{2k+1})\right|<\epsilon$$ whenever $n\ge m_\epsilon$. There is also an $m_\epsilon'\in\Bbb N$ such that $|a_k|<\epsilon$ whenever $n\ge m_\epsilon'$.
Now consider a partial sum $\sum_{k=1}^na_k$. Suppose first that $n$ is odd; say $n=2m+1$. Then $$\sum_{k=1}^na_k=a_1+\sum_{k=1}^m(a_{2k}+a_{2k+1})\;,$$ so if $m\ge m_\epsilon$ we have
$$\left|a_1+L-\sum_{k=1}^na_k\right|=\left|L-\sum_{k=1}^m(a_{2k}+a_{2k+1})\right|<\epsilon\;.$$
Now suppose that $n$ is even; say $n=2m$. Then $$\sum_{k=1}^na_k=a_1+\sum_{k=1}^{m-1}(a_{2k}+a_{2k+1})+a_{2m}\;,$$ so if $m-1\ge m_\epsilon$ and $n\ge m_\epsilon'$ we have
$$\begin{align*}
\left|a_1+L-\sum_{k=1}^na_k\right|&=\left|L-\sum_{k=1}^{m-1}(a_{2k}+a_{2k+1})-a_n\right|\\
&\le\left|L-\sum_{k=1}^{m-1}(a_{2k}+a_{2k+1})\right|+|a_n|\\
&<\epsilon+\epsilon\\
&=2\epsilon\;.
\end{align*}$$
In all cases, therefore, if $n\ge\max\{m_\epsilon,m_\epsilon'\}$, then
$$\left|a_1+L-\sum_{k=1}^na_k\right|<2\epsilon\;,$$
and it follows that $$\sum_{k\ge 1}a_k=a_1+L\;.$$
A: Let $\varepsilon>0$. Then there exists $n_0$ such that, for all $m>n$,
$$
\left|\sum_{k=n_0}^ma_{2k}+a_{2k+1}\right|<\varepsilon.
$$
There also exists $n_1$ such that $|a_k|<\varepsilon$ if $k\geq n_1$. Let $n=\max\{n_0,n_1\}$. 
Then, for any $m>2n$ and a proper $c\in\{0,1\}$,
$$
\left|\sum_{k=2n}^ma_k\right|=\left|\left(\sum_{k=n}^{\lfloor m/2\rfloor} a_{2k}+a_{2k+1}\right) - c\,a_{2\lfloor m/2\rfloor+1}\right|\leq\left|\sum_{k=n}^{\lfloor m/2\rfloor} a_{2k}+a_{2k+1}\right| +\left| \,a_{2\lfloor m/2\rfloor+1}\right|<2\varepsilon.
$$
So the tails of the series $\sum a_k$ go to zero, and so the series is convergent. 
