If subsequences are convergent, then prove that the sequence itself is convergent. Or by counter example, the sequence is divergent.

Question

Let $(a_n)^{\infty }_{n=1} $ be a sequence in a metric space such that the subsequences $(a_{2n})^{\infty }_{n=1}, (a_{2n+1})^{\infty }_{n=1}, (a_{3n})^{\infty }_{n=1}$ converge. Prove that the sequence $(a_n)^{\infty }_{n=1}$ converges.

Are we going to use the Cauchy Criterion that: " If a sequence $(x_n)$ converges then it satisfies the Cauchy’s criterion: for $\epsilon> 0$, there exists N such that $|x_n − x_m| < \epsilon $ for all $n,m \ge N$. Or something else?

Also, how can we give an example of divergent sequence $(a_n)^{\infty }_{n=1}$ such that the subsequences $(a_{2n})^{\infty }_{n=1}, (a_{2n+1})^{\infty }_{n=1}, (a_{7n})^{\infty }_{n=1}$ converge.

Can you help me?

Answer 1

We need to prove by limit definition that if

• $a_{2n}\to L_1$
• $a_{2n+1} \to L_2$

and

• $a_{3n} \to L$

then necessarly $L_1=L_2=L$ and then $a_n \to L$.

The key point is that $a_{3n}$ is a subsequence which values are alternately in $a_{2n}$ and $a_{2n+1}$.

Answer 2

You can show that $(a_{2n})^{\infty }_{n=1}$ and $(a_{2n+1})^{\infty }_{n=1}$ have to converge to the same point. Let $A=\lim_{n \to \infty} a_{2n}$ and $B=\lim_{n \to \infty} a_{2n+1}$. Assume by contradiction that $A\neq B$. Let $d = |A-B|$.

For a large enough N, we have $|a_{2n}-a_{2n+1}|>d/2 \ \forall n\geq N $. This means that $(a_{3n})^{\infty }_{n=1}$ can't converge, because it shares terms with both other sequences.

This means that $A=B$ and, because $(a_{2n})^{\infty }_{n=1}$ and $(a_{2n+1})^{\infty }_{n=1}$ define the whole sequence, that $(a_{n})^{\infty }_{n=1}$ converges.