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Even and odd subsequences of 1/(k+1) converge to 0

Example

Let (xk) be the null sequence of The sequence 1/(k+1) converges to 0, let n:N→N be the even strictly increasing index map constructed in FALSE: every bounded sequence converges, and let mj:=σ(nj) be its successor, the odd index map. Then the subsequences xnj and xmj both converge to 0 (Subsequences inherit the limit): removing infinitely many indices from this already convergent sequence, even all odd ones, preserves its inherited limit. A convergent subsequence of a divergent sequence can exist, so no converse is asserted, and by A strictly increasing index map satisfies nk≥k the index maps satisfy nj≥j and mj≥j, so the subsequences are genuine thinnings of the sequence.

Facts & Assumptions

Given: The null sequence xk:=(k+1)−1 (Sequences of reals: bounded, eventually, frequently, tails, subsequences), the index map n with n0=0 and nσ(j)=σ(σ(nj)) constructed in FALSE: every bounded sequence converges, and the successor map mj:=σ(nj).

[F1]

(xk) converges to 0 (The sequence 1/(k+1) converges to 0).

[F2]

The map n satisfies n0=0, nσ(j)=σ(σ(nj)) for every j, and n is strictly increasing (FALSE: every bounded sequence converges).

[F3]

Subsequences inherit limits: if (zk) converges to z and p:N→N is strictly increasing, then (zpj) converges to z (Subsequences inherit the limit).

[F4]

Growth of index maps: if p is strictly increasing then pj≥j for every j (A strictly increasing index map satisfies nk≥k).

[F5]

Consecutive comparisons suffice for strict increase: if pi<pσ(i) for every i, then p is strictly increasing (A strictly increasing index map satisfies nk≥k).

[F6]

Strictly increasing means pi<pj whenever i<j (Sequences of reals: bounded, eventually, frequently, tails, subsequences).

[F7]

Discreteness: for naturals a,b, a<b  ⟺  σ(a)≤b (Discreteness: σ(n) is the immediate successor).

Verification

technique · direct
1.1givenF1F2

(xk) converges to 0 and n is strictly increasing.

1.2givenF2F5F7

For every j one has mj=σ(nj)=nj+1, so m is strictly increasing: from nj<nσ(j) and [F7] we get mj=σ(nj)≤nσ(j)<σ(nσ(j))=mσ(j), that is mj<mσ(j) at every j, so [F5] applies.

2.1step 1.1F3

The subsequence along n converges to 0: [F3] applied to step 1.1 gives xnj→0.

2.2step 1.1step 1.2F3

The subsequence along m converges to 0: [F3] applied to step 1.1 and the strict increase of step 1.2 gives xmj→0.

3.1step 1.1step 1.2step 2.1step 2.2F4F6∎

Both index maps grow at least like the identity: nj≥j by [F4] applied to the strict increase of n in step 1.1, and mj≥j by [F4] applied to the strict increase of m in step 1.2. Hence the even and odd subsequences of [F6] are genuine thinnings of (xk) and both converge to the inherited limit 0.

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