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Let $h_k$ be the hitting probability of the right endpoint before the left endpoint, starting at $k$. A surviving vase corresponds to that right-endpoint event. Conditioning on the first step of the simple symmetric random walk givesThe Markov property makes the remaining probability depend only on the position after that step. These absorbing endpoint values distinguish the event of interest from its complement.

Each waiting time has the geometric distribution on $1,2,\dots$ with success probability $1/2$: a wait of length $n$ consists of $n-1$ failures followed by the prescribed success, so its probability is $2^{-n}$. HenceDisjoint successive blocks of independent tosses give independent waiting times. More explicitly, any specified list of block lengths determines exactly one allowed sequence of heads and tails and has probability $2^{-{\sum }_j(h_j+t_j)}$, the product of the individual block probabilities.

There are $2r$ waiting times. Multiplication of their probability generating functions and addition of their expectations giveTo recover its probability mass function, expand by the negative binomial series:The coefficient of $s^n$ is thereforeand it is zero below $2r$. Equivalently there are $\left(\genfrac{}{}{0ex}{}{n-1}{2r-1}\right)$ compositions of $n$ into $2r$ positive block lengths, each with probability $2^{-n}$. This is the negative binomial distribution for a sum of geometric waiting times, even though the desired outcome alternates between blocks.

For $r=1$,Since $\mathbb{E}{Y}_1=4$, Markov's inequality gives $\mathbb{P}(Y_1\ge 5)\le 4/5$, and indeed $5/16<4/5$.

For any nonnegative random variable $X$ with finite expectation and any $a>0$, Markov's inequality statesIndeed $X\ge a{1}_{\{X\ge a\}}$, and taking expectations proves the bound. Nonnegativity is essential.

First, for $0<x<1$, the derivative of $h(x)=\log (1-x)+x$ is $-x/(1-x)<0$, while $h(0)=0$. Hence $\log (1-x)<-x$.

Choose $N>c$ large enough that the given ratio bound holds for every $n\ge N$. Positivity permits taking logarithms, and the preceding inequality givesSumming, and using the integral test for convergence from part (b), gives for $n>N$Thus $a_n\le a_N{N}^c{n}^{-c}$. Since $c>1$, the P-series $\sum n^{-c}$ converges, so the comparison test for series provesThe finitely many initial terms do not affect convergence. This power-law comparison from a refined ratio bound supplies information when the ordinary ratio test can be inconclusive.

The function $1/x$ is decreasing and positive. Therefore on $[k,k+1]$ it is at most $1/k$, and the Riemann integral givesSumming from $k=1$ to $n-1$ yieldsFor $n=1$ both sides are zero; for $n>1$ the inequality is actually strict. The same integral test for convergence gives ${\sum }_{k=N}^{n-1}1/k\ge \log (n/N)$.

For a positive-term series, the ratio test says that if $a_{n+1}/a_n\to L<1$, the series converges; if the ratio tends to $L>1$, including positive infinity, it diverges. More generally $\mathrm{limsup}{a}_{n+1}/a_n<1$ suffices for convergence and $\mathrm{liminf}{a}_{n+1}/a_n>1$ for divergence. A limiting ratio of one gives no conclusion.

The P-series $\sum n^{-2}$ converges and the harmonic series $\sum n^{-1}$ diverges. Both have strictly decreasing positive terms, and their successive ratios, $(n/(n+1){)}^2$ and $n/(n+1)$, tend to one. These supply the two requested possibilities.

For a real antisymmetric matrix, $A^T=-A$, so $(A^2{)}^T=A^2$. The previous result makes every eigenvalue $\eta$ of $A^2$ real, with a real nonzero eigenvector $v$. Moreover,Thus every eigenvalue of $A^2$ is real and nonpositive.

Take real unit eigenvectors $u,w$ of $A^2$ for the respective distinct nonzero eigenvalues $-{\lambda }^2,-{\mu }^2$, where the real parameters $\lambda ,\mu$ are nonzero. DefineThe displayed quadratic identity gives $|u^{\prime }|=|w^{\prime }|=1$. Also $u\cdot Au=0$ and $w\cdot Aw=0$, because a real antisymmetric matrix has zero quadratic form. Thus $u\perp u^{\prime }$ and $w\perp w^{\prime }$. Since $A$ commutes with $A^2$, $u^{\prime }$ lies in the same eigenspace of $A^2$ as $u$, and $w^{\prime }$ in the same one as $w$. Distinct eigenspaces of the real symmetric matrix $A^2$ are orthogonal, so all four vectors are orthonormal. Finally,Thus on each of the two perpendicular planes, $A$ acts as a scaled quarter-turn. This constructs the required vectors, rather than merely counting dimensions.

Use the eigenvector basis $v_1=(-1,0,1{)}^T$, $v_2=(1,1,0{)}^T$, $v_3=(1,1,1{)}^T$. The forcing has decomposition $b=-v_1+3v_3$. If $x=c_1{v}_1+c_2{v}_2+c_3{v}_3$, the linear system reduces toFor $\lambda =0$, all coefficients are determined, givingFor $\lambda =1$, the first equation would say $0=-1$, so there are no solutions. For $\lambda =2$, $c_1=1$, $c_3=3$, and $c_2$ is arbitrary. Thus all solutions areThese outcomes illustrate respectively an invertible matrix, incompatible forcing outside its image of a linear map, and a compatible forcing with a nontrivial kernel of a linear map.

The equation is solvable exactly when $b\in \mathrm{im}A$, the image of a linear map, equivalently when the rank of a matrix satisfies $\mathrm{rank}A=\mathrm{rank}[A\mid b]$. If it is solvable and $x_0$ is one solution, every solution is preciselyIndeed differences of solutions lie in the kernel of a linear map, and adding a kernel vector preserves the equation. Hence there are no solutions when $b\notin \mathrm{im}A$; one solution when $A$ is invertible; and infinitely many solutions when $b\in \mathrm{im}A$ and $A$ is singular. In the last case the rank-nullity theorem gives a nonzero kernel vector $v$, and $x_0+tv$ gives distinct solutions for every real $t$. This proves that a finite number greater than one is impossible.

Expanding the characteristic polynomial givesThus the eigenvalues are $1,2,3$. Solving the corresponding homogeneous linear systems givesEvery nonzero vector in the displayed eigenspace is an eigenvector. Direct multiplication verifies the three eigenvalue equations. The distinct eigenvalues imply these eigenvectors are linearly independent and form a basis of ${\mathbb{R}}^3$.

For the specified unit vector,For a cyclic matrix with first row $(a,b,c)$, its determinant is $a^3+b^3+c^3-3abc$. Here it givesTherefore $Ax=Cx$ has only the zero solution if $\alpha \ne 1$. At $\alpha =1$, $A=I$ and $A-C$ has kernel consisting of the equal-coordinate vectors. The complete answer isGeometrically, $C^{T}C=I$, $\det C=1$ and $Cn=n$, so $C$ is a rotation matrix about the $n$ axis. Its matrix trace is two, giving $\cos \theta =(\mathrm{tr}C-1)/2=1/2$; its skew part selects angle $-\pi /3$ about the oriented axis $n$. It has no fixed nonzero perpendicular vector. Meanwhile $A$ preserves perpendicular components and scales the axial component by $\alpha$. Comparing these components first forces $x_{\perp }=0$, and then $(\alpha -1)x_{\parallel }=0$, exactly matching the determinant calculation, including the noninvertible case $\alpha =0$.

Expanding the formula from part (a) in components gives $y_i=x_i+(\alpha -1)n_i{n}_j{x}_j$, with repeated indices summed. Hence the matrix of a linear map and its inverse, when $\alpha \ne 0$, have entriesHere ${\delta }_{ij}$ is the Kronecker delta. To check the matrix inverse, write $P=n{n}^T$; $P^2=P$ because $|n|=1$. Thus $A=(I-P)+\alpha P$ and $B=(I-P)+{\alpha }^{-1}P$, whose product is the identity matrix.

At $\alpha =0$, $\Phi$ is the orthogonal projection onto the plane perpendicular to $n$. Its image of a linear map and kernel of a linear map areEvery perpendicular vector is unchanged and every parallel vector is annihilated. The image plane is perpendicular to the kernel axis by the definition of $n^{\perp }$; their dimensions two and one also agree with the rank-nullity theorem.

The vector triple product identity gives $n\times (x\times n)=x-(n\cdot x)n$. Decompose $x=x_{\perp }+x_{\parallel }$, where $x_{\parallel }=(n\cdot x)n$ and $x_{\perp }\cdot n=0$. ThenThe linear map acts as the identity on $n^{\perp }$ and as multiplication by $\alpha$ along $n$. Consequently it is invertible exactly when $\alpha \ne 0$, withApplying $\Phi$ to this expression recovers the perpendicular and parallel components of $y$ separately.

The original PDF has coefficient $c_n=n^{n^{1/3}}$. Its $n$th root isThe Cauchy-Hadamard theorem therefore gives $\boxed{R=1}$. Again the power series converges absolutely inside the unit circle, while on that circle its terms do not tend to zero. The converted TeX loses the outer exponent and writes $n^{1/3}$; the calculation here uses the printed coefficient.

For the power series coefficient $c_n=n^2$, the Cauchy-Hadamard theorem givesThus the radius of convergence is $\boxed{R=1}$. The power series converges absolutely for $|z|<1$ and diverges for $|z|>1$. On $|z|=1$ its terms have modulus $n^2$, so they do not tend to zero and the series diverges there as well.

The floor function has even value on indices congruent to zero or one modulo four, and odd value on indices congruent to two or three. Accordingly the displayed sequence has terms $2+1/n$ in the first two residue classes and $-1/n$ in the other two. In particular,Every term is within $1/n$ of the set $\{0,2\}$. If a subsequence converged to $c\notin \{0,2\}$, its eventual distance from that set would be bounded below by a positive constant, contradicting this bound. The complete set of subsequential limits is $\{0,2\}$.