# Convergence of $\sum_{n=1} a_n$ implies convergence of $\sum_{n=1} \left( a_{2n} + a_{2n+1} \right)$

I need to show that if the series $\sum_{n=1} a_n$ converges then the series $\sum_{n=1} \left( a_{2n} + a_{2n+1} \right)$ must also converge. Does it suffice to argue that since the two sums differ only by a single term $a_1$ and $a_n \in \mathbb{R}$ for all $n$ (since the first series converges), then the second must also converge? That is, since

$$\sum_{n=1} a_n = a_1 + a_2 + \cdots$$

and

$$\sum_{n=1} \left( a_{2n} + a_{2n+1} \right) = (a_2 + a_3) + (a_4 + a_5) + \cdots$$

we can write

$$\sum_{n=1} a_n = a_1 + \sum_{n=1} \left( a_{2n} + a_{2n+1} \right)$$

and so clearly if the left sum converges, the right must as well.

Does this seem like a sufficiently rigorous argument for an upper-division real analysis course? My class is using "Elementary Analysis: The Theory of Calculus," by the way.

• just write that convergence of $\sum u_n$ is equivalent to the existence of $\lim \sum_{k=1}^n u_k$. – mookid Mar 16 '14 at 23:10

Hint: The sequence of partial sums of $$\sum_{n=1}^\infty a_{2n}+a_{2n+1}$$ is a subsequence of the sequence of partial sums of $$\sum_{n=2}^\infty a_n$$
• But what if you consider $(a_n) = (10000, 1/2, 1/4, 1/8, \dots)$ – Rahul Shankar Mar 17 '14 at 1:57
• Neither of the sums above contain $a_1$, so the partial sums of the top series are $$(3/4,15/16,63/64,\dots)$$ and the partial sums of the bottom series are $$(1/2,\color{#C00000}{3/4},7/8,\color{#C00000}{15/16},31/32,\color{#C00000}{63/64},\dots)$$ So you see the partial sums of the first are a subsequence of the partial sums of the second. – robjohn Mar 17 '14 at 7:15