2
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ 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. $\endgroup$ – layman Nov 28 '15 at 21:18
  • 1
    $\begingroup$ Ok, but uses more machinery than necessary, one can proceed directly via $\epsilon$-$N$. $\endgroup$ – André Nicolas Nov 28 '15 at 21:24
  • $\begingroup$ Could you provide such a direct proof? $\endgroup$ – luka5z Nov 28 '15 at 21:26
2
$\begingroup$

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 $|b_k-g|=|b_{2n-1}-g|=|a_n-g|<\varepsilon$.

It looks too technical when written down, it is much easier to just think about it.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.