If $(a_{2n} -a_n)$ is convergent to zero then $(a_n)$ is convergent . if $(a_n)$ is convergent then $(a_{2n}-a_n)$ is convergent to zero

1.If $(a_{2n} -a_n)$ is convergent to zero then $(a_n)$ is convergent. True/False?

2.If $(a_n)$ is convergent then $(a_{2n}-a_n)$ is convergent. True/False?

I solved $2$ this way - let $\lim\limits_{n\to\infty }a_n =L$ and $(a_{2n})$ is a subsequence of $(a_n)$ then they must converge to the same limit , therefore $\lim\limits_{n\to\infty }a_{2n} =L$ so we get that $\lim\limits_{n\to\infty }a_{2n}- a_n =L-L=0$ so the statement is True

for the first part I know that the statement is not true but I cannot find a counterexample or a general way to show it , the practice book used this counterexample and explanation: let $a_n=\begin{cases} 1 &\text{if $n=2^k$ $k \in \Bbb N$}\\ 0&\text{otherwise}\\ \end{cases}$

if $n=2^k$ then for $k \in \Bbb N$ we get $2n=2^{k+1}$ therefore $a_{2n}=a_n=1$ and $a_{2n}-a_n=0$

if $n\not=2^k$ then $a_{2n}=a_n=0$ and $a_{2n}-a_n=0$ meaning for all $n$ we get $a_{2n}-a_n=0$ so $\lim\limits_{n\to\infty }a_{2n}- a_n =0$ but $(a_n)$ is not convergent

$(a_{2n-1})$ is a subsequence of $(a_n)$ in the odd indexes so $(a_{2n-1})=0$ and $\lim\limits_{n\to\infty }a_{2n-1}=0$. the sequence $n_k =2^k$ is a sequence that is stricly increasing of natural indexes because for all $k$ we get $2^k$ is also a natural number and $n_{k+1}=2^{k+1}=2\cdot2^k <2^k =n_k$ , we get that $(a_{n_k})=(a_{2^k})$ is a subsequence of $(a_n)$ but all of its elements are the same $\lim\limits_{n\to\infty }a_{2^k}=\lim\limits_{n\to\infty }1=1$ but this subsequence has a different limit which means that $(a_n)$ is not convergent and the statement is false

my question is if there is another way to show this? this counterexample seems too complicated , I understand the explanations but I would not have figured to use $a_n=\begin{cases} 1 &\text{if $n=2^k$ $k \in \Bbb N$}\\ 0&\text{otherwise}\\ \end{cases}$ .. is there another counterexample? or a general way to show this other than the way they did in the book?

thank you

Define the odd part of a positive integer to be its greatest odd factor (you can get it by dividing the largest possible power of $2$ out of the integer). For instance, $48 = 2^4\cdot 3$, so the odd part of $48$ is $3$.

For any integer $n$, the odd part of $2n$ is equal to the odd part of $n$. Also the odd part of any odd number is the odd number itself; so odd parts take on arbitrarily large values.

If you define your sequence by $a_n$ is the odd part of $n$, you get a divergent sequence for which $a_{2n}-a_n=0$ for all $n$.

A much simpler counterexample is given by $$a_n = \log(\log(n))$$