Proving convergence of the sequence when its subsequences converge

Can someone give me a hint on how to approach the following proof? Show that in general, a sequence $y_n$ converges if and only if the subsequences $y_{2n−1}$ and $y_{2n}$ both converge and $\lim y_{2n−1} = \lim y_{2n}$.

Intuitively, I understand that $y_{2n−1}$ represents odd terms of $y_n$ and $y_{2n}$ - even terms. Also, I know that there is a lemma stating that for the subsequence to be convergent, the limit of its subsequences has to be the same. Otherwise, it will be divergent. Since $y_{2n}$ converges then there exists a real number $N_1$, $n>N_1$ such that $|y_{2n} - l_1|< \epsilon$. Similarly, since $y_{2n−1}$ converges, then there exists a real number $N_2$, $n>N_2$ such that $|y_{2n−1} - l_2| < \epsilon$.

I guess I first need to show that the given subsequences converge to the same limit and then from that I could use the argument from the lemma.

Thanks!

• You have to assume that $l_1 =l_2$ above, otherwise you could just have sequence $01010101010101$, where the odd and even terms converge to $l_1= 0$ and $l_2 = 1$ respectively, but the sequence does not converge. Hence, work under the assumption $l_1 = l_2$. – астон вілла олоф мэллбэрг Nov 4 '16 at 0:26
• The $(\Rightarrow)$ should be clear enough. For the $(\Leftarrow)$, let $(y_{2n})$ and $(y_{2n-1})$ both converges to $L$. To show that $(y_n)$ converges to $L$, we want to make $|y_n-L|$ as small as we want. We can always do that by considering wether $n$ is odd or even. – user9077 Nov 4 '16 at 0:47
• Thank you for your tips! So, can I prove it the following way then: For (⇒) we are given that y_n converges and so by the lemma y_2n and y_2n-1 will converge as well since there are subsequences of y_n. For (⇐), we have that lim y_2n−1 = lim y_2n = L and so there exists a real number N, n>N such that |y_2n−1 - L| < ϵ and |y_2n - L| < ϵ. Sequence y_n consists of both odd and even terms and so if n is even, y_n --> L and if n is odd - y_n --> L as well. – IVU Nov 4 '16 at 8:02