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For the process to stop on toss $n$, the last toss must be the $k$th head, while the first $n-1$ tosses must contain exactly $k-1$ heads. Hence the negative binomial distribution gives

Conditioned on exactly $k$ heads, every $k$-element set of head positions is equally likely. There are $\left(\genfrac{}{}{0ex}{}{n}{k}\right)$ such sets. A run containing all $k$ heads can begin at any of the positionsand each beginning determines one admissible set. Thus, for $k\ge 1$,

The number of heads has the binomial distribution. Choosing the $k$ head positions gives

When $r$ red marbles remain, the next draw reduces their number with probability $r/n$. The waiting time $W_r$ for that reduction has a geometric distribution on $\{1,2,\dots \}$ and thereforeThe successive waiting times are independent, andUsing the product rule for probability-generating functions,

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

For $0\le s\le 1$, independence givesTherefore

The probability generating function isDifferentiating $n$ times,At $s=0$, only the term $m=n$ remains. Hence

A Cauchy sequence is a sequence $(x_n)$ such that for every $\epsilon >0$ there is $N$ for whichThe general principle of convergence, or completeness of the real numbers, states that a real sequence converges if and only if it is Cauchy.

For the final claim, let $m=\lfloor n/2\rfloor$. Since $(a_n)$ is decreasing and positive,Because the series $\sum a_k$ converges, its tails tend to zero. Also $n-m\ge n/2$, soThe squeeze theorem gives

First suppose $x_n\to L>0$. Continuity of the natural logarithm gives $\log x_n\to \log L$. Part (a), applied to this sequence, yieldsApplying the continuous exponential function,If $L=0$, then for every $\epsilon >0$ all sufficiently late $x_k$ are below $\epsilon$. Splitting off the fixed initial product shows that the limsup of the geometric means is at most $\epsilon$; hence it is zero. This proves the geometric mean of a sequence result in all cases.

Now supposeThe telescoping product isThereforeand so

Suppose $a_n\to L$. Given $\epsilon >0$, choose $N$ such that $|a_k-L|<\epsilon /2$ for $k>N$. ThenThe first term tends to zero because its numerator is fixed, and the second is at most $\epsilon /2$. Thus the Cesàro mean tends to $L$.

The converse is false. For $a_n=(-1{)}^n$, the Cesaro means tend to $0$, while the original sequence alternates between $-1$ and $1$ and does not converge.

No. Consider the two orthogonal projectionsTheir product isFor $x=(-1,2{)}^T$,Thus $PQ$ is not a positive semidefinite matrix. It is also not Hermitian; products of orthogonal projections need not remain orthogonal projections unless the factors commute.

Put $H=A-B$. This is a nonzero Hermitian matrix. By the finite-dimensional spectral theorem, it has a unit eigenvector $v$ with a nonzero real eigenvalue $lambda$. Define the rank-one orthogonal projection matrixThe cyclic property of the matrix trace givesHence this $P$ distinguishes $A$ and $B$ through the required traces.

SetThen $v=v_0+v_1$, and idempotence givesMoreover,because both $P$ and $I-P$ are Hermitian. This is the orthogonal decomposition into the kernel and image of the projection.

For an orthogonal projection matrix $P=P^{†}=P^2$,Thus $P$ is a positive semidefinite matrix.

Let $Av=\lambda v$ for a nonzero eigenvector $v$ of the Hermitian matrix $A$. Hermitian symmetry of the inner product givesSince $lVertv{\Vert }^2>0$, $lambda=\overline{\lambda }$, so every eigenvalue is real.

The relationsimply $RM=M{R}^{-1}$. Moving every occurrence of $M$ to the left and reducing exponents therefore puts every word into one of the normal formsExplicitly, with $\theta =2\pi /n$,andThe $n$ rotations are distinct. The $n$ reflected matrices are also distinct, and no reflected matrix equals a rotation because their determinants are $-1$ and $1$, respectively. Hence there are exactlymatrices, forming the dihedral group.

Because the sphere passes through the origin and has centre $P$, its radius is $|P|$. A point $X$ lies on it exactly whenor equivalentlyWith $P=\alpha A+\beta B+\gamma C$, the three required equations are the Gram matrix systemFor the specified vectors this becomesThusand the centre is