Given the sequence $\{x_n\}$, prove $\{x_{2n}\}$ and $\{x_{2n-1}\}$ converge to the same limit Let $\{a_n\}_{n=1}^{\infty}$ be a decreasing sequence of positive numbers with limit 0. I define a new sequence $\{x_n\}_{n=1}^{\infty}$ as follows:
\begin{align*}
    x_1 &= a_1 \\
    \forall n \in \mathbb{Z}^{+}, x_{n+1} &= x_n + (-1)^{n}a_{n+1}
\end{align*}
I have to prove that $\{x_{2n}\}_{n=1}^{\infty}$ and $\{x_{2n-1}\}_{n=1}^{\infty}$ are both convergent to the same limit $L$.
I have tried writing out the first few terms of the sequence:
\begin{align*}
    x_1 &= a_1 \\
x_2 &= a_1 - a_2 \\
x_3 &= a_1 - a_2 + a_3 \\
x_4 &= a_1 -a_2 + a_3 - a_4
\end{align*}
So $x_{2n} = x_{2n-1} - a_{2n}$. But I am stuck where to proceed from here. How should I show that $\{x_{2n}\}_{n=1}^{\infty}$ and $\{x_{2n-1}\}_{n=1}^{\infty}$ are both convergent to the same limit using the provided fact that $\{a_n\}_{n=1}^{\infty}$ has limit 0 and decreasing?
 A: First show that $x_{2n}$ has a limit. You can do so by observing that through telescoping: $x_{2n}-x_1 = \sum_{k=2}^{2n}(-1)^ka_k$. The right side converges by Dirichlet's test. Now use $x_{2n}-a_{2n}=x_{2n-1}$ consider the limit of the left side.
A: Look at the limit of $\{x_{2n+1} - x_{2n}\}$. If $\{x_{2n+1} - x_{2n}\} \to 0$ and $\{x_{2n}\}$,$\{x_{2n+1}\}$ are both convergent, then you can conclude they have the same limit. You can use Monotone Convergence to prove $\{x_{2n}\}$ and $\{x_{2n+1}\}$ are convergent.
A: HINT: Use the fact that $x_{2n}=x_{2n-1}-a_{2n}$ for each $n\in\Bbb Z^+$ to show that
$$\lim_{n\to\infty}x_{2n}=\lim_{n\to\infty}x_{2n-1}-\lim_{n\to\infty}a_{2n}\,,$$
and use that to show that if either of the sequences $\langle x_{2n}:n\in\Bbb Z^+\rangle$ and $\langle x_{2n-1}:n\in\Bbb Z^+\rangle$ converges, then both converge to the same limit.
The alternating series test tells you that $\sum_{n\ge 1}(-1)^{n+1}a_n$ converges, and $x_n$ is the $n$-th partial sum of this series, so ... ?
