The Real Line

The supremum is 1, which is not a member. Compact sets always attain sup and inf.

(c) fails: A = [0,1], B = [0,2] ∪ {3} … more simply A = {0, 2}, B = {0, 3} gives A ∩ B = {0} with sup 0, while min{2,3} = 2.

$n^2\sin (1/n)$ ~ n, n cos(1/n) ~ n and n/log n all tend to $+\infty$, so each eventually exceeds any x. n $e^{-n}$ ≤ 1/e is bounded, so it fails for x > 1/e.

Write $f(x)=1/2-(x-1/2{)}^2$. Fixed points are ±1/2. For |a − 1/2| < 1 the distance $d_n=1/2-x_n$ satisfies $d_n$₊${}_1=d_n^2$, which tends to 0, so $x_n\to 1/2$ for all a ∈ (−1/2, 3/2); a = 0 is included.

$x^2-9x+18\le 0\iff x\in [3,6];x^2-7x+12\le 0\iff x\in [3,4]$. So U = [3, 4] and inf U = 3.

(1) Even and odd subsequences with a common limit ⇒ convergent. (2) {$x_{6n}$} lies in both {$x_{2n}$} and {$x_{3n}$}, so those limits agree; {$x_{3(2n}$₊${}_{1)}$} lies in both {$x_{2n}$₊${}_1$} and {$x_{3n}$}, so all three limits agree ⇒ convergent. (3) Fails: $x_n=1$ if n is prime, 0 otherwise — every {$x_{kn}$} is eventually 0 but $x_n$ diverges. (4) Fails: $x_n=\sqrt{n}$.

Induction: $a_n<2\Rightarrow a_n$₊${}_1=\sqrt{2+a_n}<2$, and $a_n$₊${}_1>a_n\iff a_n^2-a_n-2<0\iff a_n<2$. Limit solves $L^2=2+L,L=2$.

(b) fails: $a_n=2$ for odd n, 1 for even n — roots → 1, ratios alternate 2, ½.

{$nmod2\pi$} is dense in $[0,2\pi )$ since $2\pi$ is irrational, so sin n is dense in [−1, 1].

(a), (b) standard. (c) is the sub-subsequence criterion — true. (d) false: $a_n=(-1{)}^n$ has Cesàro means → 0 but diverges.

Write $n^{1/n}$ = $e^{(\log n)/n}$ and (log n)/n → 0, so the limit is $e^0=1$. Alternatively set $a_n=1+h_n$ and expand $(1+h_n{)}^n\ge C(n,2)h_n^2$ to force $h_n\to 0$.

(1) The partial sums of $\sum (a_n$₊${}_1-a_n)$ form a telescoping series that converges absolutely, so {$a_n$} converges. (2) Fails: $a_n=\sqrt{n}$ has |$a_n$₊${}_1-a_n$| → 0 but is unbounded; the partial sums of the harmonic series are a bounded counterexample. (3) True, though the subsequence hypothesis is redundant — a Cauchy sequence in $\mathbb{R}$ converges anyway. (4) Fails: $a_n=(-1{)}^n$ is bounded and not Cauchy.

All the inner arguments converge: $(n\pi +(-1{)}^{n}2e)/(2n)\to \pi /2$ and $(n{\pi }^2+(-1{)}^n{e}^2)/(7n)\to {\pi }^2/7$. So the limits in (2) and (4) exist; in (1) the limit is $\exp (\cos \pi /2)=1$, not > 1; in (3) the limit is $\exp (\sin \pi /2)=e<\pi$. Official key: 3.

Even terms → 1 from above, odd terms → −1 from below.

Even $n:(1+1/n{)}^n\to e$. Odd $n:(1-1/n{)}^n\to 1/e$.

The sequence is 6-periodic with values 1, ½, −½, −1, −½, ½; the largest recurring value is 1.

(b) fails for $a_n=(-1{)}^n,b_n=(-1{)}^n$⁺${}^1$: LHS 0, RHS 2.

Even-indexed terms decrease to 1, odd-indexed terms increase to −1, and every subsequence eventually follows one of these two tracks, so the subsequential limit set is exactly {−1, 1}; limsup is its largest element 1, liminf its smallest, −1. The sequence diverges because limsup ≠ liminf.

$\sqrt{n+1}-\sqrt{n}=1/(\sqrt{n+1}+\sqrt{n})$ decreases to 0, so the alternating series converges (Leibniz). The absolute series behaves like $\sum 1/(2\sqrt{n})$, which diverges. Hence conditionally convergent — option 3 is the most precise true statement (official key: 3).

(a): the non-zero terms are $(-1{)}^k/\sqrt{2k+1}$, an alternating series with terms decreasing to 0 (Leibniz$).(b):\log (1+1/n^2)$ ~ $1/n^2$, comparison with a convergent p-series.

$a_n$₊${}_1/a_n=(n/(n+1){)}^n\to 1/e<1$.

Terms do not tend to 0 (|$a_n$| → 1).

(a) diverges by condensation $(\sum 1/(k\log 2)).(b)$ converges by condensation $(\sum 1/(k^2{\log }^{2}2)).(c)$ Leibniz. (d) $n^{1/n}$ → 1 so terms ~ 1/n — diverges.

(a) fails: $(-1{)}^n/\sqrt{n.}(b)a_n\to 0$ so eventually $a_n^2\le a_n.(c)$ fails unless absolutely convergent (Riemann).

Cauchy condensation sends $\sum 1/(n(\log n{)}^p)$ to a constant multiple of $\sum 1/k^p$: divergent for p = 1, convergent for p = 2. The third is a p-series with p = 1/2 ≤ 1, and the fourth compares with the harmonic series since $n/(n^2+1)$ ~ 1/n.