Prove series is convergent based on the convergence of the sum

Question

I need to prove:

If $\sum(a_n + b_n)$ is convergent and $\displaystyle{\lim_{n \to \infty}a_n = 0}$ then the series $$ a_1 + b_1 + a_2 + b_2 + \cdots $$ is convergent.

I was trying to use Cauchy's convergence test but I am feeling like I am missing something and I will be happy if someone can confirm my proof.

Proof: We need to show that for every $\varepsilon > 0$ there is a natural number $N$ such that:

$$|a_{n+1}+b_{n+1}+\cdots +a_{n+p}+b_{n+p} +a_{n+p+1}|<\varepsilon $$

holds for all $n > N$ and all $p \ge 1$. That's because we know the series $\sum(a_n + b_n)$ is convergent and therefore by Cauchy's convergence test for every $\varepsilon > 0$ there is a natural number $N$ such that:

$$|a_{n+1}+b_{n+1}+\cdots +a_{n+p}+b_{n+p}|<\varepsilon $$

holds for all $n > N$ and all $p \geq 1$.

From what we have just mentioned we can deduce that for every $\varepsilon/2 > 0$ there is a natural number $N_1$ such that:

$$|a_{n+1}+b_{n+1}+\cdots +a_{n+p}+b_{n+p}|<\frac{\varepsilon}{2} $$

holds for all $n > N_1$ and all $p \geq 1$.

Moreover from the fact $\lim_{n \to \infty} a_n = 0$, we can deduce that for every $\varepsilon/2 > 0$ there is a natural number $N_2$ such that:

$$|a_n| < \frac{\varepsilon}{2} \qquad\text{for all } n > N_2.$$

Let us define $N_3 = \max\{ N_1, N_2 \} $, then for all $n > N_3$ and for all $p \geq 1$:

\begin{align*} &|a_{n+1}+b_{n+1}+\cdots +a_{n+p}+b_{n+p} +a_{n+p+1}| \\ &\leq |a_{n+1}+b_{n+1}+\cdots +a_{n+p}+b_{n+p}| + |a_{n+p+1}| \\ &< \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon \end{align*}

I will be grateful for any comments for my proof, many thanks.

Answer 1

You don't need Cauchy criterion here.

Let

$$ c_n = \left\{ \begin{array}{ll} a_{(n+1)/2} & \mbox{if } n \text{ is odd} \\ b_{n/2} & \mbox{if } n \text{ is even} \end{array} \right. $$

and let $$C_n = \sum_{k=1}^n c_k$$

The question is to prove that $(C_n)_{n \in \mathbb{N}}$ converges.

Let $\displaystyle{S_n = \sum_{k=1}^n (a_k + b_k)}$. By assumption, the sequence $(S_n)_{n \in \mathbb{N}} $ converges to a limit $L$.

For every $n \in \mathbb{N}$, one has $$C_{2n} = S_n$$ $$C_{2n+1} = S_n + a_{n+1}$$

Becasue $a_n \longrightarrow 0$, then both the subsequences $(C_{2n})_{n \in \mathbb{N}}$ and $(C_{2n+1})_{n \in \mathbb{N}}$ converges to the same limit ($L$), which implies directly that $$\boxed{(C_n)_{n \in \mathbb{N}} \text{ converges.}}$$