Dtyjlui

We are given the sequence $a_n$ and it's subsequences, $a_{2n}$, $a_{2n+1}$ and $a_{3n}$, which are each convergent(and each converge to the same limit). I know that we can use $a_{2n}$ and $a_{2n+1}$ to show that $a_n$ is convergent , but can one deduce the same (that $a_n$ is convergent) with just $a_{2n}$ and $a_{3n}$ ? I know that method is similar to $a_{2n}$ and $a_{2n+1}$ but $2n$ & $3n$ do not account for all the terms.

I know that method is similar to $a_{2n} and $a_{2n+1} but $2_n$ & $3_n$ do not account for all the terms.

ANd that makes ALL the difference! If there are terms unaccounted for, those terms can do anything they want!

The point is that all $a_k$ that $k$ is either of the form $2n$ or $2n +1$ so $a_k$ in one of the sequences or the other. And if they both converge to the same limit then then whole sequence converges to the same limit.

And if you had any set of sequences $a_{k_i}, a_{j_i}, a_{m_i}$ and if the $k_i, j_i, m_i$ partitioned all the positive integers, or partitioned all but a finite number of the the positive integers so that at some point any $a_n$ must belong to one of the sequences everything would be fine.

But if there are always terms that are not in either sequence, it need not converge. $2n$ and $3n$ in no way cover all possible positive integers so we can conclude nothing from $a_{2n}$ and $a_{3n}$ converging.

Example let $a_k$ be defined as $a_k = 0$ if $2|k$ or if $3|k$ but let $a_k = k$ if neither $2$ nor $3$ divide $k$.

Then all $a_{2n} = 0$ and all $a_{3n} = 0$ so $a_{2n} to 0$ and $a_{3n}to 0$ but clearly $a_n notto 0$. (Because there are infinitely many $n$ not divisible by $2$ or by $3$ and those terms do not converge to anything.)

======

I suspect the question actually was if $a_{2n} to d$ and $a_{3n} to c$ and $a_{2n+1} to b$ then prove $a_n$ converges and determine what it converges to.

1) As $a_{6n} subset a_{2n}$ and $a_{6n} subset a_{3n}$ then $a_{6n}to d$ and $a_{6n} to c$ and $d=c$.

2) As $a_{6n + 3} subset a_{3n}$ and $a_{6n+3} subset a_{2n+1}$ then $a_{6n+3} to c=d$ and $a_{6n+3}to b$ and $b = c =d$.

3) As $a_{2n+1}$ and $a_{2n}$ completely partition $a_n$, then all $a_k$ are in one sequence or the other and those the sequence will converge to $b=c=d$.

The comment from Andreas Blass contains the answer. The only thing one need to answer your question is a counter-example. Take for instance the series $(u_n)$ such that:

• for all prime number $p>3$, $u_p = 1$
  


• for all other integer $k$, $u_k = 0$
  

The Euclid's theorem asserts that there are an infinity of prime numbers, therefore the series can not converge even though any terms of $u_{2n}$ or $u_{3n}$ will be $0$.

I think the correct question is : $(a_{2n})$, $(a_{2n+1})$ and $(a_{3n})$ are convergent, but you don't have the additional information that the limits are the same.

Now prove that $(a_n)$ is convergent.

The idea behind the proof of $ a_2n $ and $ a_(2n+1) $ being sufficient to conclude about the limit of $ a_n $ is that sequences of indices $ 2n $ and $ (2n+1) $ partition the family of indices $ I_n $, thus complete "the puzzle" if you wish. ie. they generate the family of indices $ I_n $, a subgroup of $ mathbb{Z}^+ $.

So, the procedure is without alternative; and the question about limit can be reduced to prove that **$<2n+1,3n>$ is also a generator for $ mathbb{Z}^+ $. To generate means that, there exists $ k_1,k_2 in mathbb{Z} $ such that:

$$ (1) iff k_1 (2n+1) + k_2 (3n) = N $$

where $ N $ is just any integer possible.

Observe that,

$$(1) iff n (2 k_1) + k_1 + n (3 k_2) = N $$

$$(1) iff n (2k_1 + 3k_2) + k_1 = N $$

We introduce: $ k_3 = 2k_1 + 3k_2 $:

$$(1) iff n k_3 + k_1 = N $$

In fact, we conclude that, for any $ k_3 $: $ <2n+1,3n> $ generates the quotient group $ mathbb{Z}/mathbb{Z}_(k_3) $. And, as we have no criteria over $k_3$, these two subsequences above are not sufficient to conclude anything about the main sequence's limit.