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Since $a_j=S_j-S_{j-1}$, shifting an index yields the summation by parts identityThe sum on the last line is empty when $m=n$, so the identity also covers that case.

The series with a bounded monotone multiplier converges. Here is a proof that also permits conditional convergence. Choose $M>0$ with $|b_j|\le M$. The Cauchy sequence criterion for the convergent series $\sum a_j$ ensures that, for any $\eta >0$ and sufficiently large $m$, all the partial tails $T_j={\sum }_{k=m}^j{a}_k$ satisfy $|T_j|<\eta$. Apply summation by parts to these partial tails:Since the multiplier is a monotone sequence, ${\sum }_{j=m}^{n-1}|b_j-b_{j+1}|=|b_m-b_n|$. ConsequentlyTaking $\eta =\epsilon /(3M)$ proves that the weighted partial sums form a Cauchy sequence, and hence converge. This is a special case of bounded-variation multipliers of a convergent series.

The series $\sum n^{1/n}{a}_n$ also converges. The multiplier need only be monotone eventually, since a finite initial segment cannot affect convergence. For $h(x)=\log x/x$,Thus $n^{1/n}=\exp (h(n))$ decreases for $n\ge 3$; it is bounded and tends to $1$, because $\log n/n\to 0$. Apply the proved multiplier result to the tail beginning at $3$. The fact that the first few values are not monotone causes no difficulty.

A complex power series with radius of convergence $R$ converges absolutely at every $z$ with $|z|<R$ and diverges at every $z$ with $|z|>R$. This makes no general assertion at $|z|=R$: boundary points must be examined separately. The series always converges at $z=0$. If $R=0$, no nonzero point lies in the convergence disk; if $R=\infty$, the series converges absolutely throughout the complex plane.

If $p$ is the zero polynomial, every coefficient vanishes, so $R=\infty$. Otherwise let $d$ be its degree and $c\ne 0$ its leading coefficient. Since $p(n)=c{n}^d(1+o(1))$,The ratio test therefore proves absolute convergence when $|z|<1$. For $|z|>1$, the terms $p(n)z^n$ do not tend to zero, so the series diverges. Thus the polynomial-coefficient power series has $R=1$ for every nonzero polynomial. In fact, it also diverges at every $|z|=1$, because $|p(n)z^n|=|p(n)|$ does not tend to zero, whether $d=0$ or $d>0$.

Denote the four displayed matrices, from left to right, by $T_1,T_2,T_3,T_4$. An orthogonal matrix has an orthonormal set of columns, equivalently $T^{T}T=I$. Direct column dot products give $T_j^T{T}_j=I$ for $j=1,2,4$. The first column of $T_3$ has squared Euclidean norm $(1+6+1)/6=4/3$, so the third matrix is the unique nonorthogonal matrix.

For $T_1$, the determinant is $1$. A nonidentity orthogonal transformation of ${\mathbb{R}}^3$ with determinant $1$ is a rotation: it has a fixed axis and restricts to a plane rotation on its orthogonal complement. Consequently $T_1$ is the rotation.

For $T_2$, calculation givesA plane reflection fixes its entire plane, and therefore has eigenvalue $1$. Since $T_2-I$ is invertible, $T_2$ is not a plane reflection. Thus $T_2$ is the combination of a rotation and a reflection. More explicitly, $(1,-2,0{)}^T$ is an eigenvector with eigenvalue $-1$. On the perpendicular plane, $T_2$ is a rotation with $\cos \varphi =2/3$, because $\mathrm{tr}{T}_2=-1+2\cos \varphi =1/3$. This gives a three-dimensional improper orthogonal transformation.

For $T_4$, set $n=(1,2,2{)}^T/3$. ThenThis reflection matrix reverses the normal $n$ and fixes every vector perpendicular to $n$. Therefore $T_4$ is reflection in the plane $x+2y+2z=0$. In particular, $\det T_4=-1$ and $\det (T_4-I)=0$. Finally $T_3$ represents none of the three listed orthogonal transformations, since each preserves the Euclidean norm.