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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?

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    $\begingroup$ 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. $\endgroup$ – Bonbon May 9 at 13:55
  • $\begingroup$ 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 $\endgroup$ – Calvin Khor May 9 at 14:31
  • $\begingroup$ @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. $\endgroup$ – Ashish K May 9 at 14:39
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    $\begingroup$ @AshishK the two sequences have to '(eventually) cover all the points'. For instance its not enough to check $a_{3n}$ and $a_{3n+1}$. $\endgroup$ – Calvin Khor May 9 at 14:41
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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.

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  • $\begingroup$ 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 $A_k + B_k$. $\endgroup$ – Ashish K May 9 at 14:00
  • $\begingroup$ Isn't $A_k + B_k$ different from $S_k$ (as defined in my answer)? $\endgroup$ – Ashish K May 9 at 14:03
  • $\begingroup$ @AshishK what do you mean by different? $\endgroup$ – Darius May 9 at 14:10
  • $\begingroup$ 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) $\endgroup$ – Ashish K May 9 at 14:18
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    $\begingroup$ 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 $\alpha+\beta$. $\endgroup$ – Calvin Khor May 9 at 14:23

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