If $\sum_{n=1}^{\infty} a_n$ and $\sum_{n=1}^{\infty} b_n$ both converge with sums $\alpha$ and $\beta$, then show that the series $a_1+b_1+a_2+b_2+a_3+b_3+\ldots$ converges with sum $\alpha + \beta$.
Here's my attempt at a proof-
Let $(S_n)$ be the sequence of partial sums of the series $a_1+b_1+a_2+b_2+a_3+b_3+\cdots$. Let $(T_n)$ and $(U_n)$ be the sequence of partial sums of the series $\sum_{n=1}^{\infty} a_n$ and $\sum_{n=1}^{\infty} b_n$ respectively. Then we can define $(S_n)$ as follows: $$S_n = \begin{cases} T_{n/2}+U_{n/2} & \text{if n is even} \\ T_{\frac{n+1}{2}}+U_{\frac{n-1}{2}} & \text{if n is odd} \\ \end{cases} $$
Let $\varepsilon > 0$ be given. Then there are $N_1, N_2 \in \mathbb{N}$ such that $|T_{k+1}-\alpha| < \frac{\varepsilon}{2}$ for all $k\ge N_1$ and $|U_{k}-\beta| < \frac{\varepsilon}{2}$ for all $k\ge N_2$.
Let $N=\max \{ 2N_1+1, 2N_2+1 \}$. Then we will show that $|S_n - (\alpha +\beta)|< \varepsilon$ for all $n\ge N$.
Assume $n\ge N$. If $n$ is odd, then $n=2k+1$ for some $k \in \mathbb{N}$. Hence, $k\ge N_1$ and $k\ge N_2$ and it follows that $$\begin{align} |S_n - (\alpha + \beta)| &= |S_{2k+1} - (\alpha + \beta)| \\ &= |(T_{k+1} + U_{k})-(\alpha + \beta)| \\ & < |T_{k+1} - \alpha| + |U_{k}- \beta| \\ & < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon \end{align}$$
Likewise we can do the same with $n$ being even part.
Is my proof correct? Is there a way to complete the proof by avoiding the definition of a sequence?
Your proof looks good to me!
An easier method would be to use the Algebraic Limit Theorem as follows:
Let $A_k = \sum_{n=1}^{k} a_n$ and $B_k = \sum_{n=1}^{k} b_n$. Since both sequences $(A_k: k\in\mathbb{N})$ and $(B_k: k\in\mathbb{N})$ converge, their respective limits $\alpha, \beta$ exists. Then by the Algebraic Limit Theorem,
$$\lim\limits_{n\to\infty}(A_k + B_k) = \lim\limits_{n\to\infty}(A_k) + \lim\limits_{n\to\infty}(B_k) = \alpha+\beta$$
Note that the proof for the Algebraic Limit Theorem is actually pretty similar to what you're doing!
Edit: This is not a complete proof as pointed out by Calvin Khor.