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For the geometric distribution on $\{1,2,\dots \}$,Summing the geometric series,

The special binomial theoremfollows by mathematical induction. Multiplication of the formula for $n$ by $1+t$ and collection of the coefficient of $t^k$ uses Pascal's identity from part (a).

Termwise integration from $0$ to $1$ givesReplacing $t$ by $-t$ gives

Finally,Writing $y=1-x$ and using the finite geometric series,turns this into

For each permitted part $i\le e$, its multiplicity contributes the geometric series $1+x^i+x^{2i}+\cdots$. Thus the generating function isThe conjugate partition bijects diagrams with largest part at most $e$ and diagrams with at most $e$ rows, so the same series counts partitions into at most $e$ parts.

Now try only $k_i=2^i$. Since $n^{\ast }=2^N=\Theta (1/\theta )$, the total work through trial $N$ is the geometric seriesFor $i<N$, the success probabilities scale asTheir sum is $O(1)$, so the product of their failure probabilities stays bounded away from zero: there is a constant probability of reaching the scale $k_N=n^{\ast }$. At that trial, the defining property of $n^{\ast }$ givesso almost all surviving runs stop there. ThereforeThe geometric amplitude-amplification schedule is asymptotically better than the sequential schedule's $\Theta ((n^{\ast }{)}^{4/3})$ calls and matches the usual Grover scaling up to a constant factor.

Split the transform into periods and translate each interval:Summing the geometric series yields

WriteThe stated iteration is $x_{n+1}=B{x}_n+{\alpha }^{-1}{A}^{\ast }y$, so repeated substitution and the geometric series identity giveFor the singular component $v_i$, put $r_i=1-{\sigma }_i^2/\alpha$. Since the initial Tikhonov solution has coefficient ${\sigma }_i(y,u_i)/({\sigma }_i^2+\alpha )$, one obtainsTherefore the requested spectral filter isThus equation (3) reads $x_n={\sum }_i{g}_{\alpha ,n}({\sigma }_i)(y,u_i)v_i$. Convergence of this stationary iteration requires $|1-{\sigma }_i^2/\alpha |<1$ on the nonzero spectrum, for example $\alpha >\Vert A{\Vert }^2/2$.

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

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

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 $1/\sqrt{2}$.

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.

First the expected value of the absorption time is finite. From any interior state, consecutive leftward moves lead to absorption within at most $n$ steps, with probability at least $p=3^{-n}$. By independence of the subsequent steps, the probability of surviving $jn$ steps is at most $(1-p{)}^j$. Summing this geometric series bounds the expected value of the absorption time by $n/p$.

Now apply first-step analysis. Starting from $k$ in the interior, one step is always used. Conditional on that step, the remaining expected values are $e_{k+1}$, $e_k$, and $e_{k-1}$, each with probability $1/3$. The law of total expectation therefore givesThe term $1$ remains even when the particle stays at $k$, because a step has elapsed. Rearranging gives

Define the radius of convergence of the power series bywith $R=\infty$ when this set is unbounded. It is nonempty because the power series converges at zero.

If $|z|<R$, the definition of supremum supplies a point $w$ of convergence with $|z|<|w|$. The terms $a_n{w}^n$ of a convergent series tend to zero and are bounded, say $|a_n{w}^n|\le M$. ThereforeThe right-hand side has a convergent geometric series, since $|z|/|w|<1$. The comparison test proves absolute convergence at $z$.

If $|z|>R$ and the power series converged at $z$, then $|z|$ would belong to the set whose supremum is $R$, an immediate contradiction. ThusNo assertion about the boundary $|z|=R$ follows from this argument. If $R=0$ the first assertion is vacuous; if $R=\infty$ the second is vacuous.

Split the power series into even and odd indices:whenever the two series converge absolutely. If $|z|<1$, the even part is a convergent geometric series, while the odd part is bounded termwise in absolute value by ${\sum }_j|z{|}^{2j+1}$, also a convergent geometric series. Hence there is absolute convergence throughout $|z|<1$.

If $|z|>1$, the odd terms have absolute values $|z{|}^{2j+1}/(2j+1{)}^2$. Their successive ratio tends to $|z{|}^2>1$, so these terms grow eventually and do not tend to zero. The term test for divergence therefore forces divergence of the original power series. Consequently its radius of convergence is

The finite geometric series identity gives, for each $|x|<1$,For every fixed $n$, the absolute value of this error tends to infinity as $x\to 1^{-}$. Thusand there is no uniform convergence on $(-1,1)$.

Using the geometric series, the probability generating function isDifferentiation gives

We give the explicit point $\boxed{x=1/4}$. Repeatedly choose the left remaining third and then the right remaining third. After $2m$ choices, the resulting closed interval is $[L_m,R_m]$, whereThese formulas follow from a finite geometric series. They show $L_m<1/4<R_m$, and that both endpoints tend to $1/4$. The point $1/4$ is also strictly inside each intermediate left third: its distance from $L_m$ is one quarter of the current length. It therefore never enters any assigned closed middle third or its endpoints, so $1/4\in C$.

Endpoint values in the Cantor function construction obey the same left/right choices, with halving in place of thirding. ThusAll these endpoints are in the domain of the partial function: $L_m$ is the right endpoint of a middle third assigned at round $2m$, and $R_m$ is the left endpoint of a middle third assigned at round $2m-1$.

Given $\epsilon >0$, choose $m$ so that $(2/3)4^{-m}<\epsilon$, and put $\delta =9^{-m}/4$. If $t\notin C$ and $|t-1/4|<\delta$, then $L_m<t<R_m$. The assumed monotonicity givesConsequently $|f(t)-1/3|<\epsilon$. This proves directly, on the stated partial domain,There are defined points arbitrarily close to $1/4$, for example $L_m$ and $R_m$, so the limit of a function is not vacuous. No continuous extension has been assumed. The residual $C$ here omits the endpoints that the question assigns values to; it is the usual Cantor set with those assigned endpoints removed.

The binomial theorem gives $\left(\genfrac{}{}{0ex}{}{2n}{n}\right)\le {\sum }_{k=0}^{2n}\left(\genfrac{}{}{0ex}{}{2n}{k}\right)=4^n$. ThereforeThe geometric series with ratio $4/9$ converges. By the comparison test, the given series is absolutely convergent.

For a power series $\sum a_n{z}^n$, its radius of convergence $R\in [0,\infty ]$ separates absolute convergence for $|z|<R$ from divergence for $|z|>R$. One may define $R=\sup \{r\ge 0:\sum |a_n|r^n<\infty \}$, allowing $R=\infty$.

To justify convergence inside the radius, choose $|z|<r<R$ with $\sum |a_n|r^n<\infty$. In particular $|a_n|r^n\le M$ for some $M$, soThe comparison test with a geometric series proves absolute convergence. Conversely, convergence at a point of modulus $\rho$ makes $a_n{z}^n$ bounded there and gives absolute convergence at all smaller moduli by the same estimate. Consequently a convergent point of modulus greater than $R$ would contradict the definition of $R$.

For $0<R<\infty$, the real domain of convergence is one of $(-R,R)$, $[-R,R)$, $(-R,R]$, or $[-R,R]$. At $R=1$ these are realized respectively byThe endpoint assertions follow from the harmonic series, alternating series test, and p-series. Replacing $z$ by $z/R$ realizes every finite positive radius. The degenerate domain $\{0\}$ occurs for $\sum n!z^n$, while the entire real line occurs for $\sum z^n/n!$. Their radii are zero and infinity by the ratio test.