subsequence

Subsequence

Let $(n_1,n_2,n_3,\dots )$ be an be an increasing sequence of natural numbers and $(a_n{)}_{n\in \mathbb{N}}$ be a sequence. Then we call

$$(a_{n_k}{)}_{k\in \mathbb{N}}=(a_{n_k}{)}_{k=1}^{\infty }=(a_{n_1},a_{n_2},\dots )$$

a subsequence of $(a_n{)}_{n\in \mathbb{N}}$.

$k$ indexes the subsequence (1st pick, 2nd pick, …), $n_k$ says where each pick lives in the original. The requirement $n_1<n_2<n_3<\dots$ means you pick in order, but can skip terms.

Even-indexed subsequence of $b_n=(-1{)}^n\left(1-\frac{1}{n}\right)$

The full sequence: $0,\frac{1}{2},-\frac{2}{3},\frac{3}{4},-\frac{4}{5},\frac{5}{6},\dots$

Pick $n_k=2k$ (even indices): $b_2,b_4,b_6,\cdots =\frac{1}{2},\frac{3}{4},\frac{5}{6},\cdots \to 1$

Pick $n_k=2k-1$ (odd indices): $b_1,b_3,b_5,\cdots =0,-\frac{2}{3},-\frac{4}{5},\cdots \to -1$

The full sequence diverges (it jumps), but each subsequence converges.

Subsequence of primes

Let $a_n=1+\frac{1}{n}$ and pick only prime indices: $n_k=2,3,5,7,11,\dots$

Subsequence: $\frac{3}{2},\frac{4}{3},\frac{6}{5},\frac{8}{7},\frac{12}{11},\cdots \to 1$

Same limit as the full sequence. This always happens: if $(a_n)\to L$, then every subsequence also $\to L$ (there’s no way to “escape” by picking terms, since all terms eventually stay near $L$).