# Convergent series…is this a correct method?

Here is the question: Let $(a_n)$ (for $n\in\mathbb{N}$) be a sequence of real number and let $S_k$ denote the partial sums of the series $$\sum_{n=1}^\infty a_n$$ Prove that if $\lim_{k\to \infty}$$S_{2k}=\ \lim_{k\to \infty}$$S_{2k+1}$=$L$, then $$\sum_{n=1}^\infty a_n=L$$

Here's my attempt:

$S_{2k}$ is the sequence of even partial sums such as: $(S_2, S_4, S_6, ...)$ and $S_{2k+1}$is the sequence of odd partial sums $(S_3, S_5, S_7,....)$ . There is a theorem in our book that states subsequences of a convergent sequence converge to the same limit as the original sequence. So can I say that since these are both converging subsequences of $S_k$ that converge to the same limit L, then $S_k$ converges to L?

Thanks.

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I don't think you can use that theorem like that. If you know that $S_k$ converges to $L$, then you can conclude that both $S_{2k}$ and $S_{2k+1}$ converge to $L$, but not the other way around. – Javier Nov 11 '12 at 0:15
Wouldn't that be the case in an "if, then" statement? – Alti Nov 11 '12 at 0:18

What you know is that, given $\varepsilon>0$, there exists $2n>N_1$ such that $n>N_1$ implies $|S_{2n}-L|<\varepsilon$; and that there exists $N_2$ such that $2n+1n>N_2$ implies $|S_{2n+1}-L|<\varepsilon$.
Now let $N=\max\{N_1,N_2\}$. Then, if $k>N$, then either $k=2n>N_1$, or $k=2k+1>N_2$; in both cases, $|S_k-L|<\varepsilon$. So the sequence converges to $L$.
If you do, you are already assuming that the original series converges. If you do not assume that the original series converges, then you can show that it actually does by using the definition of a convergent series, i.e. that the sequence of partial sums converges to $L$.