Assume that $(a_n)_{n \in \mathbb{N}}$ tends to $0$, show that the series $\sum a_n$ and $\sum (a_n+a_{n+1})$ either both converge or both diverge 
Assume that $(a_n)_{n \in \mathbb{N}}$ tends to zero. Show that the series $\sum_{n=1}^\infty a_n$ and $\sum_{n=1}^\infty (a_n+a_{n+1})$ either both converge or both diverge.

My textbook notice the following:
$$\sum_{k=1}^n (a_k+a_{k+1})-2\sum_{k=1}^n a_k=a_{n+1}-a_1 \to -a_1$$
However, I tried another proof. My proof: assume by contradiction that $\sum_{n=1}^\infty a_n$ converges and $\sum_{n=1}^\infty (a_n+a_{n+1})$ diverges. Since $a_{n+1}$ is obtained by $a_n$ with a finite translation of index, the series $\sum_{n=1}^\infty a_{n+1}$ converges as well and so for the theorem of the sum of two convergent series I can add the two series like this:
$$\sum_{n=1}^\infty a_n+\sum_{n=1}^\infty a_{n+1}=\sum_{n=1}^\infty (a_n+a_{n+1})$$
But the left hand side is convergent because sum of two convergent series and the right and side is divergent; a contradiction.
Similarly, assume by contradiction that $\sum_{n=1}^\infty a_n$ diverges and $\sum_{n=1}^\infty (a_n+a_{n+1})$ converges. Since $a_{n+1}$ is obtained by $a_n$ with a finite translation of index, the series $\sum_{n=1}^\infty a_{n+1}$ diverges as well (and for $\sum_{n=1}^\infty (a_n+a_{n+1})$ there aren't indeterminate forms $\infty-\infty$ or $-\infty+\infty$ because they diverge with the same sign) and so for the theorem of the sum of two divergent series I can add the two series like this:
$$\sum_{n=1}^\infty a_n+\sum_{n=1}^\infty a_{n+1}=\sum_{n=1}^\infty (a_n+a_{n+1})$$
But the left hand side is divergent because sum of two divergent series (one obtained by translation from the other) and the right and side is convergent; again a contradiction.
Is this proof correct? I didn't use explicitly that $a_n \to 0$ as $n \to \infty$, even if that is necessary to assume the convergence of $\sum_{a_n}$. So I believe that my proof is wrong somewhere. Maybe I am not considering the cases when the series can be indeterminate? Or is it impossible that the series are indeterminate and my proof is valid?
 A: The error lies in the assertion “the left hand side is divergent because sum of two divergent series”. If two series $\sum_{n=1}^\infty a_n$ and $\sum_{n=1}^\infty b_n$ are divergent, then the expression $\sum_{n=1}^\infty a_n+\sum_{n=1}^\infty b_n$ makes no sense. In particular, it neither converges nor diverges.
A: First suppose $\sum a_n$ converges. Then the limit
$\lim_{n\to \infty} \sum_{k=1}^n a_k$ exists (in our normed space and thus in $\mathbb{R}$ or $\mathbb{C}$ in particular). Let us put
$$
l := \lim_{n \to \infty} \sum_{k=1}^n a_k. \tag{ Definition 1} 
$$
For each $n \in \mathbb{N}$, we can write
$$
\sum_{k=1}^n \left( a_k + a_{k+1} \right) = -a_1 + a_{n+1} + 2 \sum_{k=1}^n a_k,
$$
therefore we obtain
$$
\begin{align} 
\lim_{n \to \infty} \sum_{k=1}^n \left( a_k + a_{k+1} \right) &=  -a_1 +  \lim_{n\to\infty} a_{n+1}  + 2 \lim_{n \to \infty} \sum_{k=1}^n a_k \\ 
&= -a_1 + 0 + 2l \\ 
& \qquad \qquad \mbox{[ using (Definition 1) above ]} \\ 
&= -a_1 + 2l,
\end{align} 
$$
which shows that the series $\sum \left( a_n + a_{n+1} \right)$ converges.
Conversely, suppose that the series $\sum \left( a_n + a_{n+1} \right)$ converges. Then $\lim_{n \to \infty} \sum_{k=1}^n \left( a_k + a_{k+1} \right)$ exists. Let us put
$$
m := \lim_{n \to \infty} \sum_{k=1}^n \left( a_k + a_{k+1} \right). \tag{Definition 2} 
$$
For each $n \in \mathbb{N}$, we can write
$$
\sum_{k=1}^n a_k = \frac{1}{2} a_1 - \frac{1}{2} a_{n+1} + \frac{1}{2} \sum_{k=1}^n \left( a_k + a_{k+1} \right).
$$
So we obtain
$$
\begin{align} 
\lim_{n \to \infty} \sum_{k=1}^n a_k &= \frac{1}{2} a_1 - \frac{1}{2} \lim_{n \to \infty} a_{n+1} + \frac{1}{2} \lim_{n \to \infty} \sum_{k=1}^n \left( a_k + a_{k+1} \right) \\ 
&= \frac{1}{2} a_1 - 0 + \frac{1}{2} m \qquad \mbox{[ using (Definition 2) above ]} \\
&= \frac{1}{2} a_1 + \frac{1}{2} m,
\end{align} 
$$
thus showing that the series $\sum a_n$ also converges.
