# Proving if $\sum_{n=1}^{\infty} a_n$ and $\sum_{n=1}^{\infty} b_n$ both converge then the series $a_1+b_1+a_2+b_2+\ldots$ converges.

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?

• The proof is right. And I think that the definition of $S_{n}$ is not so nontrivial, even avoiding the definition, you still have to do similar thing. – Bonbon May 9 '19 at 13:55
• To me, the proof is cleaner if you separate out the following result, true for any sequence $a_n$: $$a_{2n}\to a \text{ and } a_{2n+1}\to a \implies a_n \to a$$ Then for $S_{2n}$ and $S_{2n+1}$, you can appeal to the sum rule for series, which avoids $\epsilon$ management – Calvin Khor May 9 '19 at 14:31
• @CalvinKhor i recall a result stating "A sequence converges to L iff all of its subsequences converge L". I was wondering if checking for only two subsequences would be enough or not. – Ashish K May 9 '19 at 14:39
• @AshishK the two sequences have to '(eventually) cover all the points'. For instance its not enough to check $a_{3n}$ and $a_{3n+1}$. – Calvin Khor May 9 '19 at 14:41

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.

• The reason I avoided is the fact that the sequence of the partial sum of $a_1+b_1+a_2+b_2+a_3+b_3+\cdots$ is quite different from $A_k + B_k$. – Ashish K May 9 '19 at 14:00
• Isn't $A_k + B_k$ different from $S_k$ (as defined in my answer)? – Ashish K May 9 '19 at 14:03
• @AshishK what do you mean by different? – Darius May 9 '19 at 14:10
• We say that a series converges if its sequence of partial sums converges. The sequence of partial sums of $a_1+b_1+a_2+b_2+a_3+b_3+\cdots$ is not $A_k+B_k$ so I believe the two are not equivalent. Here's the problem from my text: i.imgur.com/… (the part (b) of Theorem 6.1.5 that the textbook refers to is what you did) – Ashish K May 9 '19 at 14:18
• Strictly speaking, I suppose this only proves that the sum of the first 2k terms (whose partial sums are indeed $A_k+B_k$) converges to $\alpha+\beta$. – Calvin Khor May 9 '19 at 14:23