Choose the unbounded function $\boxed{g(x)=1/x}$ on $E=(0,1/2]$. It is unbounded because $g(1/k)=k$ for all integers $k\ge 2$. For each $x\in E$, the geometric series gives the sum and its remainder after degree $N$:Since $0<x\le 1/2$,This proves uniform convergence. Although the first term $1/x$ and the sum are unbounded, the late tails have a bound tending to zero independently of $x$. Uniform convergence concerns those tails and does not require the individual functions to be bounded.

Let $S_N={\sum }_{n=1}^N{f}_n$ and let $s$ be its uniform limit. Choose a finite $B\ge 0$ with $|g(x)|\le B$ for all $x\in E$. The partial sums of the multiplied series are $g{S}_N$, andFor any $\epsilon >0$, uniform convergence supplies $N_0$ making $|S_N(x)-s(x)|<\epsilon /(B+1)$ for all $N\ge N_0$ and $x\in E$. The displayed estimate then gives an error less than $\epsilon$. Thus

Take $0\le r<R$. At the point $r$, the power series has absolute convergence, so ${\sum }_{n=0}^{\infty }|a_n|r^n<\infty$. On the whole interval $[-r,r]$,The Weierstrass M-test with $M_n=|a_n|r^n$ proves uniform convergence on $[-r,r]$. It also gives the explicit tail estimateFor $r=0$ the interval consists only of zero, and all positive-degree terms vanish. If $R=\infty$, the same argument applies to every finite $r$; if $R=0$, there is no $r\ge 0$ with $r<R$.

The Weierstrass M-test states: if functions $f_n:E\to \mathbb{R}$ or $\mathbb{C}$ satisfythen $\sum f_n$ has uniform convergence and absolute convergence on $E$. In fact $\sum |f_n|$ also has uniform convergence, since every tail is bounded by the corresponding tail of $\sum M_n$:

Write the partial sums as $S_N(x)={\sum }_{n=1}^N{f}_n(x)$. The series has uniform convergence to $s:E\to \mathbb{R}$ ifThe same threshold $N_0$ must control every point of $E$. Equivalently, the supremum error ${\sup }_{x\in E}|S_N(x)-s(x)|$ tends to zero, for nonempty $E$.

TakeBoth functions have values in $[1,\infty )$. The reverse triangle inequality gives $|g(x)-g(y)|\le |x-y|$, so $g$ is Lipschitz continuous and hence has uniform continuity.

For $x_n=n$ and $y_n=n+1/n$, the distances satisfy $|x_n-y_n|=1/n\to 0$, whereasIn particular tolerance $\epsilon =1$ defeats every proposed $\delta$: choose $n$ with $1/n<\delta$. Thus $f$ does not have uniform continuity on $\mathbb{R}$.

For any $x,y\in \mathbb{R}$, rationalizing the difference of square roots givessince $f(x),f(y)\ge 1$. Given $\epsilon >0$, use uniform continuity of $f$ with tolerance $2\epsilon$ to obtain $\delta >0$. Whenever $|x-y|<\delta$, the displayed estimate yields $|g(x)-g(y)|<\epsilon$. Thus $g$ has uniform continuity on $\mathbb{R}$. The square root function is in fact Lipschitz continuous with constant $1/2$ on $[1,\infty )$.

Suppose the continuous function $f$ did not have uniform continuity on $[a,b]$. Negating its definition gives ${\epsilon }_0>0$ such that for every $n\ge 1$ there are $x_n,y_n\in [a,b]$ withThe Bolzano-Weierstrass theorem, which states that every bounded sequence of real numbers has a convergent subsequence, gives $x_{n_k}\to p$. The inequalities $a\le x_{n_k}\le b$ imply $p\in [a,b]$. By the triangle inequality,Since continuous functions preserve convergent sequences, both $f(x_{n_k})$ and $f(y_{n_k})$ tend to $f(p)$. Their difference therefore tends to zero, contradicting its absolute value being at least ${\epsilon }_0$. This proves the Heine-Cantor theorem in the required setting: $f$ has uniform continuity on $[a,b]$.

A function $f:E\to \mathbb{R}$ is continuous on $E$ if it is continuous at every $p\in E$, meaningThe function has uniform continuity on $E$ whenFor continuity, $\delta$ may depend on both $p$ and $\epsilon$. For uniform continuity, one $\delta$ must work for every pair of points in $E$.

Use the ceiling function to choose positive integer coefficientsBecause $0<\alpha <1$, we have ${\alpha }^{-n}\ge 1$ andFor $|z|<\alpha$, the comparison test for series with the geometric series givesFor $|z|\ge \alpha$, we instead have $|a_n{z}^n|\ge (|z|/\alpha {)}^n\ge 1$, so the terms do not tend to zero. Thus the power series has radius of convergence exactly $\alpha$.

All three examples below have positive integer coefficients, so they work under either convention for whether $0\in \mathbb{N}$.

For $R=1$, take $\boxed{a_n=1}$. This is the geometric series: it converges for $|z|<1$, and its terms fail to tend to zero for $|z|\ge 1$.

For $R=0$, take $\boxed{a_n=n!}$, using the factorial and $0!=1$. For every fixed $z\ne 0$, the absolute values of successive terms have ratio $(n+1)|z|$. This eventually exceeds two, so the terms grow without bound and the series cannot converge. At $z=0$ it is finite, giving radius of convergence zero.

For $R=1/\sqrt{2}$, takeusing the floor function. The coefficients satisfyThe upper bound and the comparison test for series give absolute convergence when $\sqrt{2}|z|<1$. The lower bound prevents the terms tending to zero when $\sqrt{2}|z|\ge 1$. Hence the radius of convergence is exactly $1/\sqrt{2}$.

If $z_0=0$, there is no $z$ with $|z|<|z_0|$, so the claim is vacuous. Suppose $z_0\ne 0$. Convergence of $\sum a_n{z}_0^n$ implies $a_n{z}_0^n\to 0$, because terms of a convergent series tend to zero. In particular this is a bounded sequence, so $|a_n{z}_0^n|\le C$ for some finite $C$ and every $n$.

For $|z|<|z_0|$, put $q=|z|/|z_0|<1$. ThenThe geometric series and comparison test for series proveThus convergence at one point gives absolute convergence at every point strictly closer to the centre.

The comparison test for series states that, for $0\le u_n\le v_n$ eventually, convergence of $\sum v_n$ implies convergence of $\sum u_n$. Equivalently, divergence of $\sum u_n$ implies divergence of $\sum v_n$.

For $|z|<1$, the coefficient bound givesby the geometric series formula and the comparison test for series. The single term $a_0$ does not affect convergence. For $|z|\ge 1$,so the terms do not tend to zero. The fact that terms of a convergent series tend to zero excludes convergence there. Thus

The radius of convergence of a power series ${\sum }_{n=0}^{\infty }{a}_n{z}^n$ is the number $R\in [0,\infty ]$ such that the series has absolute convergence for $|z|<R$ and diverges for $|z|>R$. Equivalently,Use $r^0=1$, including at $r=0$. The supremum may be $+\infty$. There is no general conclusion at $|z|=R$: boundary convergence depends on the coefficients. If $R=0$, only the centre is guaranteed; if $R=\infty$, every finite $z$ lies inside the disk of convergence.

The finite geometric series identity, obtained by subtracting $t$ times the sum from the sum itself, givesFor $0<t<1$, the geometric sequence $t^{N+1}$ tends to zero by Question 2(a)(iii). Therefore

Let $T_N={\sum }_{n=1}^N|a_n|$ and $S_N={\sum }_{n=1}^N{a}_n$. If $(T_N)$ converges, it is a Cauchy sequence. Thus, for every $\epsilon >0$, some $N$ makeswhere the triangle inequality gives the bound. Therefore $(S_N)$ is a Cauchy sequence in the complex numbers. These form a complete metric space, so $(S_N)$ has a finite limit of a sequence. Equivalently, apply the Cauchy criterion for series. Hence absolute convergence implies convergence.

For a series of complex numbers, form the partial sums $S_N={\sum }_{n=1}^N{a}_n$. The series converges when these partial sums have a finite limit of a sequence $S\in \mathbb{C}$:The series has absolute convergence when ${\sum }_{n=1}^{\infty }|a_n|$ converges as a series of nonnegative real numbers. A series beginning at $n=0$ uses the same definitions with that starting index.

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

True. Choose $N$ such that $\ell /2<a_n<3\ell /2$ for every $n\ge N$. All the finitely many preceding terms are also positive. Define the positive minimum and finite maximumEvery term lies between $\eta$ and $B$, so every ratio is at most $B/\eta$. Consequently the required supremum satisfies

True. Since $a_n^n\to 2$, there is $N_0$ such that $1<a_n^n<3$ for $n\ge N_0$. Positivity then gives $a_n>1$ on this tail.

Fix $\epsilon >0$. If $a_n\ge 1+\epsilon$, Bernoulli's inequality givesFor $n>2/\epsilon$, this contradicts $a_n^n<3$. Therefore eventually $1<a_n<1+\epsilon$, proving