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The term at $k=0$ is finite. For every $k\ge 1$,The comparison P-series $\sum k^{-2}$ converges, so the comparison test for series proves that the given series converges absolutely.

By part (e) and the comparison test for series, $\sum u_n$ converges, since the P-series $\sum n^{-2}$ converges. Put $S={\sum }_{n=1}^{\infty }{u}_n$. Part (a) now provesPart (d) also shows that $(a_n)$ increases towards this finite positive limit. This proves convergence of the Stirling sequence without presupposing its limiting constant.

For identification of that constant, the independently known Stirling's formula $n!\sim \sqrt{2\pi n}(n/e{)}^n$ givesThe comparison argument above supplies the convergence proof requested by the preceding parts; the Stirling's formula identification is an additional evaluation.

For positive integers $n$, the summand is $a_n=\sin (\pi \sin (1/n))/n$. Since $0<\sin (1/n)<1$, these terms are positive. Apply the limit comparison test with the convergent p-series $\sum n^{-2}$:Each sine ratio tends to one. The limit comparison test says that two positive-term series whose term ratio tends to a finite positive number are either both convergent or both divergent. ThereforeThere is also a direct comparison test for series: $0\le \sin (\pi \sin (1/n))\le \pi \sin (1/n)\le \pi /n$, giving $0\le a_n\le \pi /n^2$.

Write a nonterminating parity series as $y=x^r{\sum }_{j\ge 0}{b}_j{x}^{2j}$, where $r=0$ or $1$. Its recurrence relation isThe ratio test applied to its nonzero successive terms has limiting absolute ratio $|x{|}^2$. Hence the series converges absolutely for $|x|<1$ and diverges for $|x|>1$, withAt $x=\pm 1$, the limiting ratio is one, so the basic ratio test alone is inconclusive. Retain the next term of the ratio instead:For large $j$ the ratio is positive. Taking logarithms and summing gives $\log |b_j|=-\frac{3}{2}\log j+O(1)$; the $O(j^{-2})$ remainders have a finite sum. Thus $|b_j|\le C{j}^{-3/2}$ eventually. Comparison with the convergent p-series proves absolute convergence at both endpoints as well:The radius of convergence one is the distance from the expansion point to the nearest singular points of a differential equation, $x=\pm 1$. Convergence of function values at those points does not make the solutions analytic there; endpoint derivatives can diverge. The terminating Chebyshev polynomial is an exception with infinite radius of convergence. For example, at $m=0$ the other solution is $\arcsin x$, whose endpoint values are finite while its first derivative diverges.

This is the synchronous product of simple symmetric random walks, rather than a walk that changes only one coordinate at a time. Independence of the three coordinate processes givesOdd-time return probabilities vanish. The p-series with exponent $3/2$ converges, so the sum of all return probabilities is finite. By the recurrence criterion by return probabilities, the origin is a transient state: the three-dimensional process is transient. From the origin its communicating class consists of integer triples whose coordinates have the same parity; translation gives the same classification for every other class.

For the symmetric simple random walk on the integer line, odd return times are impossible. A return after $2m$ steps requires exactly $m$ positive increments, soby Stirling's approximation. The resulting sum diverges by the p-series criterion. The recurrence criterion by return probabilities therefore makes zero a recurrent state. The simple random walk is an irreducible Markov chain on $\mathbb{Z}$, so the symmetric chain is recurrent.

For $p,q>0$, the positive series converges precisely when $p>1$, or $p=1$ and $q>1$. For $p>1$ use comparison with a convergent p-series; for $0<p<1$, the bound $(\log n{)}^q<n^{(1-p)/2}$ gives a lower comparison with a divergent p-series. At $p=1$, the integral test for convergence and $u=\log x$ reduce the question to $\int u^{-q}du$. In contrast, ${\sum }_{n\ge 3}[n(\log \log n{)}^r{]}^{-1}$ diverges for every finite $r>0$, since $(\log \log n{)}^r<\log n$ eventually and $\sum [n\log n{]}^{-1}$ diverges.

Suppose $a_n>0$ and, for all sufficiently large $n$, $a_{n+1}/a_n\le 1-c/n$ with fixed $c>0$. Choose an initial index $N>c$. Since $\log (1-x)\le -x$ for $0<x<1$, summing logarithmic ratios gives $\log (a_n/a_N)\le -c{\sum }_{k=N}^{n-1}1/k\le -c\log (n/N)$. Thus $a_n\le a_N{N}^c{n}^{-c}$. For $c>1$, the comparison test for series with a P-series proves convergence. This supplies a useful refinement when a limiting ratio of one makes the ordinary ratio test inconclusive.

For $p>0$, the function $x\mapsto \sin x/x^p$ is Lebesgue integrable on $(0,\infty )$ exactly when $1<p<2$. Near zero, $\sin x/x\to 1$, so absolute integrability is equivalent to that of $x^{1-p}$, which requires $p<2$. At infinity, $p>1$ gives the upper bound $x^{-p}$. For $0<p\le 1$, on each interval $[k\pi ,(k+1)\pi ]$,and the corresponding P-series diverges.

The signed improper integral nevertheless converges for every $0<p<2$: near zero the same local comparison applies, and at infinity the Dirichlet test for improper integrals uses the bounded antiderivative of sine and the decreasing factor $x^{-p}\to 0$. Thus $0<p\le 1$ gives conditional convergence, while $1<p<2$ gives absolute integrability. In particular, substituting $t=1/x$ turns ${\int }_0^1|x^{-1}\sin (1/x)|dx$ into the divergent tail ${\int }_1^{\infty }|\sin t|/tdt$.

Take independent random variables $U_n$, each with the continuous uniform distribution on $[0,1]$, and set $A_n=\{U_n\le p_n\}$. Then $\mathbb{P}(A_n)=p_n$, and the events are mutually independent events. The product probability space supplies a model for the entire infinite sequence; a finite list of simulated values is only an illustration.

For $p_n=1/(n+1{)}^2$, the p-series is summable:The Basel problem supplies the total sum, and the integral test for convergence gives the displayed tail bound. The First Borel-Cantelli lemma makes the limsup of events have probability zero. In fact, the tail bound implies that the probability of even one alarm at or after $m$ is at most $1/m$; for $m=100$, it is at most $0.01$. This bound remains valid for dependent events with the same individual probabilities.

For $p_n=1/(n+1)$, the harmonic series diverges, so the Second Borel-Cantelli lemma makes the limsup of events have probability one. There is also an exact finite-horizon calculation using independent events:For any fixed $m$, this finite product of numbers tends to zero as $M\to \infty$. Every infinite tail therefore contains an alarm almost surely, even though the individual probabilities tend to zero.

Take $d$ independent copies of the simple random walk on the integer line, each stepping left or right with equal probability at every integer time. Their product Markov chain moves all coordinates at once; each of its $2^d$ diagonal increments has probability $2^{-d}$. At odd times a return to the origin is impossible. At time $2n$, independence gives the displayed return probability, whose asymptotic equivalence follows from Stirling's approximation. The recurrence criterion by return probabilities and the p-series therefore imply recurrence for $d=1,2$ and a transient Markov chain for $d\ge 3$.

For $d>1$ the chain from the origin visits only the sublattice whose coordinates all have the same parity. This is one communicating class; restricting to it gives an irreducible Markov chain. The period of a state in a Markov chain is two here. The conclusion applies to that class, without incorrectly claiming Markov chain irreducibility on all of ${\mathbb{Z}}^d$. This walk moves diagonally; the usual axis-step simple random walk has different transition probabilities, even though the same dimensional recurrence threshold holds.

For $p>0$ and $n\ge 2$, integrating the decreasing derivative magnitude of $x^{-p}$ gives $(n-1{)}^{-p}-n^{-p}=p{\int }_{n-1}^n{x}^{-p-1}dx>p{n}^{-p-1}$. Put $p=k-1$ and sum this telescoping series to obtain the bound in the title, strict for $N>1$. The partial sums are increasing and bounded, proving convergence of the p-series when $k>1$. This gives a direct comparison test for series proof and a finite upper bound without evaluating the sum.