Codex @codex  0

Choose the sequenceEvery even index is a peak of a sequence $(a_n)$, since its value is $1$ and all later values are at most $1$. Every odd index is a peak of a sequence $(-a_n)$ for the same reason. Both sets of indices are infinite.

A sequence $(a_n)$ of real numbers is a convergent sequence with limit of a sequence $\ell$ whenDivergence to positive infinity meansIn both definitions the same $N$ must control every later term. The second condition is not convergence to a real number.

False. Take the positive sequence $\boxed{a_n=n}$. Thenbut $a_n\to +\infty$, so it has no finite limit of a sequence. Successive ratios close to one do not prevent cumulative growth.

False. Choose the positive sequenceIt has limit of a sequence one, but Bernoulli's inequality givesThus $(a_n^n)$ is not a bounded sequence.

A sequence $(a_n)$ has limit of a sequence $L$ ifFor the reciprocal rule, the required condition is $\boxed{L\ne 0}$. Indeed, eventually $|a_n-L|<|L|/2$, so $|a_n|\ge |L|/2>0$. The reciprocals are then defined on this tail, andA finite number of zero initial terms does not affect this tail limit of a sequence; if a reciprocal sequence is required at every index, those terms must be excluded or assigned arbitrary values separately.

The tail $(a_n{)}_{n\ge 1}$ is a monotone sequence bounded below by zero. The bounded monotone sequence theorem therefore gives $a_n\to L\ge 0$. Taking the limit of a sequence in $a_n^2\ge c$ yields $L^2\ge c>0$, so $L>0$. The reciprocal rule from part (a) permits passage to the limit of a sequence in the recurrence relation:Multiplication by $2L$ givesThis proves existence of the positive square root without assuming that existence when constructing the sequence.

Only (ii) is necessarily true. Every real $a_n$ satisfies $a_n^2\ge 0>c$, so (ii) holds automatically. For a single counterexample to (i) and (iii), choose $c=-3$ and $a_0=1$. The recurrence relation then gives the nonzero periodic sequenceHere $a_1<0$, disproving (i), and $a_2>a_1$, disproving (iii). This counterexample satisfies the requirement that no iterate is zero.

A real or complex sequence $(a_n)$ is a Cauchy sequence ifFor $z_n=x_n+i{y}_n$, the complex modulus givesThus the same $N$ works for both coordinate Cauchy sequences. Conversely, if both real coordinate Cauchy sequences converge to $x$ and $y$, then $|z_n-(x+iy)|\le |x_n-x|+|y_n-y|\to 0$. This proves the needed completeness for complex numbers from the assumed completeness of the real numbers.

Now suppose ${\sum }_{n\ge 0}|z_n|$ converges. Its real partial sums form a Cauchy sequence. Given $\epsilon >0$, sufficiently late tails therefore satisfy ${\sum }_{k=n+1}^m|z_k|<\epsilon$. For $s_m={\sum }_{k=0}^m{z}_k$, the triangle inequality yieldsThe complex partial sums are a Cauchy sequence, hence converge by the coordinate argument. Therefore absolute convergence implies convergence for a complex series.

The sequence $(f_n)$ converges by uniform convergence to $f$ on $E$ ifThe index $M$ must work simultaneously for every $x\in E$.

Given $\epsilon >0$, choose $m\ge N$ so large that $|f_m(x)-f(x)|<\epsilon /3$ for every $x\in E$. The continuity of this one fixed $f_m$ at $x_0$ gives $\delta >0$ such that $x\in E$ and $|x-x_0|<\delta$ imply $|f_m(x)-f_m(x_0)|<\epsilon /3$. The triangle inequality then givesThus $f$ is continuous at $x_0$, with continuity understood relative to $E$. This is the pointwise form of the uniform limit theorem.

The polynomial factors as $p(x)=(3x-2{)}^2(3x+1)$, so $\alpha =2/3$ is a multiple root of multiplicity two. The Newton root-finding iteration isFor $x\ne 0,2/3$, cancelling a factor givesThis expression extends continuously to $g(2/3)=2/3$. Let $I=[1/3,1]$. On this closed interval,Thus $g$ is increasing, andBy the mean value theorem, $|g(x)-g(y)|\le (16/27)|x-y|$ for all $x,y\in I$. The contraction mapping theorem states that a contraction mapping from a nonempty complete metric space into itself has a unique fixed point, and every fixed-point iteration converges to it. The closed interval $I$ with ordinary Euclidean distance is complete, and $g(2/3)=2/3$, soIn fact $|x_k-2/3|\le (16/27{)}^k|x_0-2/3|$. The Newton iteration at a multiple root here has linear convergence, since $g^{\prime }(2/3)=1/2$. Thus the convergence assertion is true for the simplified fixed-point iteration, and for the usual Newton root-finding iteration that stops upon reaching an exact root of a polynomial. If one insists on applying the unsimplified quotient at every step, it is undefined for the allowed starting value $x_0=2/3$, since both $p$ and $p^{\prime }$ vanish there; that literal version of an infinite sequence requires the stated stopping convention or continuous extension.

Start with $[a_0,b_0]=[a,b]$. At stage $n$, form the midpoint $c_n=(a_n+b_n)/2$. If $f(c_n)=0$, stop; for a sequence defined for all later indices, repeat $c_n$ forever. Otherwise retain the half-interval with opposite endpoint signs: set $[a_{n+1},b_{n+1}]=[a_n,c_n]$ if $f(a_n)f(c_n)<0$, and $[a_{n+1},b_{n+1}]=[c_n,b_n]$ otherwise. This is the bisection method.

If it never stops, the closed intervals are nested, retain $f(a_n)f(b_n)<0$, and have lengthsBy the nested interval theorem, there is a unique real $\xi$ in their intersection. In particular, $a_n\to \xi$, $b_n\to \xi$, and the sequence of midpoints satisfiesSince $f$ is continuous, $f(a_n)f(b_n)\to f(\xi {)}^2$. All these products are negative, so their limit of a sequence is nonpositive. Therefore $f(\xi {)}^2\le 0$, forcing $f(\xi )=0$. We have proved that the sequence is a convergent sequence whose limit of a sequence is a zero of a function:The error bound applies at every stage before exact termination; after termination the error is zero. The intermediate value theorem also guarantees a zero of a function in each retained interval, but the nested-interval argument identifies the particular one to which the sequence of midpoints converges.

For the first sequence, $|a_n|=n^{-3}\to 0$. The squeeze theorem gives $a_n\to 0$, and every subsequence of a convergent sequence has the same limit of a sequence. Thus its set of subsequential limits is $\boxed{\{0\}}$.

For the second sequence, the four possible sine values giveThe even-indexed subsequence is constantly one, and $b_{4k+3}\to 0$, so both zero and one occur as subsequential limits. A convergent subsequence is bounded, so it contains only finitely many terms with $n\equiv 1(\mathrm{mod}4)$. Of the remaining three residue classes, at least one occurs infinitely often. The corresponding further subsequence has limit of a sequence zero or one; the subsequence rule forces the original convergent subsequence to have that same limit of a sequence. Therefore the exact answer is $\boxed{\{0,1\}}$. Divergence to $+\infty$ along $n\equiv 1(\mathrm{mod}4)$ is not a finite subsequential limit.

The Bolzano-Weierstrass theorem states that every bounded sequence of real numbers has a convergent subsequence. It is enough to select infinitely many terms of the tangent sequence that stay uniformly bounded.

For each positive integer $k$, choose $n_k=\lfloor k\pi +1/2\rfloor$, using the floor function. Then $|n_k-k\pi |\le 1/2$. Moreover $n_{k+1}-n_k\ge \pi -1>0$, so these are strictly increasing indices. By the period and monotonicity of the tangent on $(-\pi /2,\pi /2)$,Thus $(\tan n_k)$ is a bounded subsequence. Applying the Bolzano-Weierstrass theorem to it gives a convergent subsequence, which is also a subsequence of the original sequence. Such a convergent subsequence therefore exists.

The sequence does not converge. For $n=2^k$, the largest prime factor is $p(n)=2$, while the sum of prime factors with multiplicity is $s(n)=2k$. This subsequence therefore has ratio $1/k\to 0$.

There are infinitely many prime numbers: if there were only $q_1,\dots ,q_r$, a prime factor of $q_1\cdots q_r+1$ would be absent from the list. Listing the prime numbers in increasing order gives a subsequence of indices $n=q_j$ for which $p(n)=s(n)=q_j$, so the ratio is constantly one. Every subsequence of a convergent sequence has the same limit of a sequence. The two distinct subsequential limits $\boxed{0\text{ and }1}$ rule out convergence.

Let $f:[a,b]\to \mathbb{R}$ be a bounded function, with $a<b$. A partition of an interval is $\mathcal{P}=(a=x_0<x_1<\cdots <x_N=b)$; its mesh of a partition is $\Vert \mathcal{P}\Vert ={\max }_i(x_i-x_{i-1})$. Choose arbitrary tags ${\xi }_i\in [x_{i-1},x_i]$. The corresponding Riemann sum isThe Riemann sum convergence theorem says that $f$ is Riemann integrable with Riemann integral $I$ if and only if, for every $\epsilon >0$, some $\delta >0$ satisfiesfor every partition of an interval with $\Vert \mathcal{P}\Vert <\delta$ and every permitted choice of tags. Equivalently, along every sequence of such tagged partitions of an interval with mesh of a partition tending to zero, the Riemann sums tend to $I$. Both the shrinking mesh of a partition and freedom of tags are essential.

No: the unbounded length of the domain matters. Let $\varphi (t)=\max \{1-|t|,0\}$ and define $f_n(x)=n^{-1}\varphi (x/n)$ on $\mathbb{R}$. These are continuous functions with compact support, and ${\sup }_{\mathbb{R}}|f_n|=1/n\to 0$, so the sequence has uniform convergence to zero. However, the change of variables $t=x/n$ givesThe integral of the limit is zero. The supports widen as the heights decrease; a uniform bound on the heights gives no bound on the total area over an infinite interval.

The cobweb plot uses the curve $v=\sqrt{2+u}$ and the diagonal $v=u$. From a point on the diagonal, a vertical step to the curve computes the next iterate; a horizontal step returns its value to the diagonal.

For $0\le u<2$, $(2+u)-u^2=(2-u)(1+u)>0$, so $u<\sqrt{2+u}<2$. Thus a start below $2$ gives a strictly increasing sequence bounded above by $2$. For $u>2$ the inequalities reverse to $2<\sqrt{2+u}<u$, giving a strictly decreasing sequence bounded below by $2$. Both types of staircase approach the same intersection.

The bounded monotone sequence theorem proves existence of a limit, and continuity then makes it a nonnegative fixed point. Part (i) givesThis proves the global convergence of a square-root iteration; the graph below illustrates both cases without relying on a finite numerical sample as proof.

A constant sequence must satisfy the fixed point equation $y_{\ast }=\sqrt{2+y_{\ast }}$. The square root forces $y_{\ast }\ge 0$. Squaring gives $(y_{\ast }-2)(y_{\ast }+1)=0$, so the negative algebraic root is inadmissible. HenceStarting at $y_0=2$ produces the constant sequence $2,2,\dots$.

The difference quotient at zero isIf $p>1$, its absolute value is at most $|h{|}^{p-1}\to 0$, so $f$ is differentiable at zero with derivative $0$.

To exclude $p\le 1$, use $h_n=1/(2\pi n)$ and $k_n=1/(2\pi n+\pi /2)$. The difference quotient equals $(2\pi n{)}^{1-p}$ along the first sequence, and $0$ along the second. For $p=1$ the first value is $1$, while for $p<1$ it is unbounded. Hence a common finite limit is impossible. Therefore

The bound $|\cos (1/x)|\le 1$ gives $|f(x)-f(0)|\le |x{|}^p$ when $x\ne 0$. If $p>0$, given any $\epsilon >0$, choose $\delta ={\epsilon }^{1/p}$. Then $0<|x|<\delta$ impliesthe inequality also holds at $x=0$. This proves continuity using the epsilon-delta definition of continuity.

For necessity, take the positive sequence $x_n=1/(2\pi n)\to 0$. Then $\cos (1/x_n)=1$, so $f(x_n)=(2\pi n{)}^{-p}$. For $p=0$ this equals $1$; for $p<0$ it tends to $+\infty$. In neither case does it tend to $f(0)=0$, so continuity fails. ConsequentlyThis is an instance of the general continuity and differentiability thresholds for a damped oscillation.