# Prove that the sequence $a_1, a, a_2, a, a_3, a,\ldots$ converges iff $a_1,a_2,a_3,\ldots$ converges

Prove that the sequence $a_1, g, a_2, g, a_3, g,\ldots$ converges to $g$ iff $a_1,a_2,a_3,\ldots$ converges to $g$.

Obviously, if $a_1, g, a_2, g, a_3, g,\ldots$ converges to $g$, then its subsequence $a_1,a_2,a_3,\ldots$ converges to $g$. On the contrary, suppose that $a_1,a_2,a_3,\ldots$ converges to $g$. Then $a_1, g, a_2, g, a_3, g,\ldots$ is a Cauchy sequence. Since it has a subsequence convergent to $g$, it must converge to $g$ itself.

Is this the correct reasoning?

• Good point regarding your comment on my (now deleted) answer. You're right, since the sequence is cauchy, and has a convergent subsequence, the entire sequence converges. I agree with your answer. – layman Nov 28 '15 at 21:18
• Ok, but uses more machinery than necessary, one can proceed directly via $\epsilon$-$N$. – André Nicolas Nov 28 '15 at 21:24
• Could you provide such a direct proof? – luka5z Nov 28 '15 at 21:26

Here is a direct proof (as suggested by @AndréNicolas in the comments) that if $a_1,a_2,a_3\cdots\to g$ then $a_1,g,a_2,g,a_3,g\cdots\to g$. Note that the latter sequence is $b_1,b_2,b_3\cdots$ , where $b_{2n}=g$ and $b_{2n-1}=a_n$. So take $\varepsilon>0$, then there is $N$ such that if $n>N$ then $|a_n-g|<\varepsilon$. Let $K=2N-1$. Then if $k>K$ and $k$ is even, $k=2n$, we have $|b_k-g|=|b_{2n}-g|=|g-g|=0<\varepsilon$, and if $k>K$ and $k$ is odd, $k=2n-1>K=2N-1$, then $n>N$, hence $|b_k-g|=|b_{2n-1}-g|=|a_n-g|<\varepsilon$.