(Proof Verification) If $\lim_{k\to\infty}(2k)a_{2k}=0$ and $\lim_{k\to\infty}(2k+1)a_{2k+1}=0$ then $\lim_{n\to\infty}na_n=0$.

Question

Suppose that $\lim_{k\to\infty}(2k)a_{2k}=\lim_{k\to\infty}(2k+1)a_{2k+1}=0$. But suppose that it's not true that $\lim_{n\to\infty}na_n=0$. Then, there are two cases for $na_n$. For first case, $na_n\to L$ for some $L \neq > 0$. If so, then we must have all of its subsequences must also converge to $L$. This contradicts the hypothesis. For the second case, where $na_n \to \infty$, for any $M\in\mathbb{R}$ we can find $n\in\mathbb{N}$ such that $na_n>M$. But this also contradicts the hypothesis because there exists $N\in\mathbb{N}$ such that for all $n > \geq N$ implies $na_n < \epsilon$ since $(2k)a_{2k}$ is one of such $n$. $\blacksquare$

I used proof by contradiction but I want to know if I can derive $\lim_{n\to\infty}na_n$ from $\lim_{k\to\infty}(2k)a_{2k}=0$ and $\lim_{k\to\infty}(2k+1)a_{2k+1}=0$, from using the definition of the limit or Cauchy.

Answer 1

Well, of course you can.

Let $\epsilon > 0$. Then, there exists $N_1, N_2$ such that $n > N_1 \implies |(2n)a_{2n}| < \epsilon$, and $n > N_2 \implies |(2n+1)a_{2n+1}| < \epsilon$.

Let $N = \max{N_1,N_2}$. Let $n > 2N$. Suppose $n$ is even, then $n = 2m$ for some $m > N$, so $m > N_1$, so $|(2m)a_{2m}| < \epsilon$, but since $n=2m$, $|na_n| < \epsilon$.

Suppose $n$ is odd, then $n = 2m+1$ for some $m > N$. Then, $m > N_2$, so $|(2m+1)a_{2m+1}| < \epsilon$, but then since $n = 2m+1$, $|n a_n| < \epsilon$.

Hence, whenever $n > 2N$, $|na_n|<\epsilon$. Hence, we are done i.e. the limit is zero.

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To give another method, let $b_n = na_n$. We know that $b_{2n}$ and $b_{2n+1}$ both converge to $0$.

Now, if $b_{n_k}$ is any subsequence of $b_n$, then $b_{n_k}$ must contain infinitely many odd or even indices $n_k$. If it contains infinitely many odd indices, then it contains a subsequence of $b_{2n+1}$, which is convergent to $0$. Similarly, if it contained infinitely many even terms.

Thus, every subsequence of $b_n$ contains a convergent subsequence which converges to $0$. It follows that $b_n \to 0$, since if not, then for some $\epsilon > 0$ there exists $n_k$ with $|b_{n_k}| > \epsilon$ for all $k$, and this sequence cannot have a convergent subsequence, contradiction.

So $b_n \to 0$. This proof also shows that if the natural numbers are broken up into finitely many ordered sequences $s_i$ such that $b_{s_i}$ converges for each $i$, then $b_n$ also converges.

Answer 2

Your proof is correct, but it is not complete. What happen is $\lim_{n \to \infty} na_n$ doesn't exists in $\mathbb{R} \cup \left\{{\infty}\right\}$?

Answer 3

Mpre generally, if $(S_i)_{i=1}^n$ are infinite sets of positive integers whose union is all the positive integers, and $\lim_{n \in S_i} na_n =0$ for all $i$ then $\lim na_n =0$.

To prove this, for any $c>0$, choose the max $N$ such that $|na_n|N$ and $n \in S_i$.

Answer 4

This is another shorter approach: you have $\lim_{n\to\infty}\frac{a_{2n}}{\frac{1}{2n}}=0$ and $\lim_{n\to\infty}\frac{a_{2n+1}}{\frac{1}{2n+1}}=0$. Then, since the subsequences of the even and odd numbers span all of $\mathbb{N}$, any other subsequence of $\mathbb{N}$ going to $\infty$ has any of its elements either in the even or in the odd subsequence. Then, $\lim_{n\to\infty}\frac{a_{n}}{\frac{1}{n}}=0$.