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Knowing the radius of convergence does not settle convergence on its boundary. The three power seriesall have radius one, but behave differently on $|z|=1$.

For the geometric series, boundary terms all have complex modulus one, violating the term test for divergence. For the second series, absolute values give the divergent harmonic series at every boundary point. At $z=1$ the series itself is harmonic and diverges. At any other point on the unit circle, its geometric partial sums obeyThe Dirichlet test with the decreasing sequence $1/n\to 0$ therefore gives conditional convergence. In particular, $z=-1$ gives the negative of the alternating harmonic series, with sum $-\ln 2$. For the third series, the convergent p-series ${\sum }_{n\ge 1}1/n^2$ bounds the absolute values at every boundary point, giving absolute convergence by the comparison test for series.

At $z=-1$, the rational function $1/(1-z)$ is well defined and equals $1/2$, yet its geometric series diverges there. A formula for a function and a particular series representation can have different domains. Likewise, applying termwise differentiation of a power series preserves the radius but need not preserve the boundary convergence.

The function is continuous for $x>0$, hence Lebesgue measurable. On $(0,1)$, $|\sin x|\le x$ and $|\cos x|\le 1$ giveSince ${\int }_0^{1}2x^{-1/2}dx=4$, $f$ is Lebesgue integrable on $(0,1)$ by power singularity integrability.

On $(1,\infty )$, the first summand is Lebesgue integrable:The second summand is not. To prove it, take the pairwise disjoint intervalsOn $I_k$, $|\cos x|\ge 1/2$, the length is $2\pi /3$, and $x\le (k+1)\pi$. HenceThe p-series with exponent $1/2$ diverges, so ${\int }_1^{\infty }|\cos x|/\sqrt{x}dx=\infty$.

If $f$ were Lebesgue integrable on this tail, subtracting its integrable first summand would make $\cos x/\sqrt{x}$ integrable as well, a contradiction. ThereforeThis uses the general fact that an integrable perturbation cannot repair nonintegrability. The conclusion concerns absolute integrability; oscillatory cancellation in an improper integral is a separate issue.

The function is continuous for $x>0$, hence Lebesgue measurable. Near zero, the bound $|\sin x|\le x$ givesThus $f$ is Lebesgue integrable on $(0,1]$ for $q<2$. The convergence is an instance of power singularity integrability.

For the tail, take the pairwise disjoint intervalsOn each $I_k$, $|\sin x|\ge 1/2$, $x\le (k+1)\pi$, and the interval length is $2\pi /3$. ThereforeSumming the nonnegative integrals over these disjoint intervals yieldsby divergence of the p-series. Hence $f$ is not Lebesgue integrable on $[1,\infty )$ for $0<q\le 1$.

The test is absolute integrability: oscillatory sign cancellation does not make the absolute-value integral finite. A conditionally convergent improper integral is a different notion. The sine power kernel integrability criterion separates these endpoint thresholds.

Near zero, $\sin x/x\to 1$, so $|\sin x|/x^{\alpha }$ is comparable to $x^{1-\alpha }$. The power integral test shows that the integral near zero is finite exactly when $\alpha <2$.

For the tail, $|\sin x|\le 1$ proves integrability when $\alpha >1$. To prove necessity, take the disjoint intervalsOn each, $\sin x\ge 1/2$, and $x\le 2\pi (k+1)$. Therefore ${\int }_{I_k}|\sin x|x^{-\alpha }dx\ge c_{\alpha }(k+1{)}^{-\alpha }$ with $c_{\alpha }>0$. The p-series diverges when $\alpha \le 1$, proving that the absolute tail integral diverges in that range.

Combining the two conditions,For $0<\alpha \le 1$, the tail improper integral converges conditionally by the Dirichlet test for improper integrals, but that does not give Lebesgue integrability.

Compare with the divergent series from part (i). The ratio of its new summand to $1/(n\log n)$ isFor completeness, $\log n={\int }_1^n{x}^{-1}dx\le {\int }_1^n{x}^{-1/2}dx=2(\sqrt{n}-1)$, so $(\log n)/n\to 0$. The limit comparison test now gives divergence. Although the exponent $1+1/n$ exceeds one for each summand, it approaches one and does not produce a convergent p-series with a fixed exponent greater than one.

The integral test for convergence applies when $a_n=h(n)$ for all sufficiently large $n$, where $h$ is nonnegative, continuous and nonincreasing on $[N,\infty )$. It states that ${\sum }_{n=N}^{\infty }{a}_n$ converges if and only if the improper integral ${\int }_N^{\infty }h(x)dx$ is finite. Its comparison bounds areNo conclusion about an arbitrary nonnegative sequence follows without such a monotone comparison function; finite initial terms do not affect convergence.

For $h(x)=x^{-\alpha }$ with $\alpha >0$, all the hypotheses hold. If $\alpha \ne 1$,whereas for $\alpha =1$ the integral is $\log R$. The first expression has a finite limit exactly when $\alpha >1$. Thus the p-series satisfies

For odd $n$, the denominator is $(n+1{)}^2$; for even $n$, it is $(n-1{)}^2$. There is no zero denominator in the printed index range. For all $n\ge 2$,The comparison test for series with the convergent p-series gives absolute convergence. In fact the consecutive pair $n=2k-1,2k$ contributes $1/(2k{)}^2+1/(2k-1{)}^2$, so the sum is the same as ${\sum }_{m=1}^{\infty }{m}^{-2}$. Positivity justifies this paired reordering; an alternating-sign appearance in the denominator does not make this an alternating series.

For $n\ge 2$, $0<n^{-2}\le [n(n-1){]}^{-1}=1/(n-1)-1/n$. The upper bound has telescoping partial sums, bounded by one. The positive partial sums of the original series are consequently bounded by two and increasing, so they converge as a monotone bounded sequence. The series converges absolutely. This also supplies the $p=2$ case of a p-series without requiring its exact sum.

For $z\ne 0$, the ratio of successive absolute terms isThe ratio test therefore gives absolute convergence for $|z|<1$ and divergence for $|z|>1$, while at zero the series is trivially convergent. Hence $R=1$. On the boundary, $|z^n/n^2|=1/n^2$, and the convergent p-series proves

The comparison test for series says that an eventual upper bound by a convergent positive series proves convergence, while an eventual lower bound by a divergent positive series proves divergence. The p-series $\sum n^{-s}$ converges exactly when $s>1$.

If $p>1$, then $(\log n{)}^q\ge 1$ for all sufficiently large $n$, soand the series converges. If $0<p<1$, choose $\eta =(1-p)/2>0$. The allowed logarithmic bound, applied with exponent $\eta /q$, gives $(\log n{)}^q<n^{\eta }$ eventually. Thusand the series diverges.

For $p=1$, use the integral test for convergence: for an eventually positive decreasing function $g$, $\sum g(n)$ and $\int g(x)dx$ either both converge or both diverge. Here $g(x)=1/[x(\log x{)}^q]$ is decreasing for $x>1$, and the substitution $u=\log x$ givesThis integral converges exactly when $q>1$. Hence the polynomial-logarithmic series convergence classification isAll other allowed positive $p,q$ give divergence.

For $a>0$ and $n\ge 2$, let $A_n(a)=\{X_n^2>a\log n\}$. Symmetry of the standard normal distribution and the Mills ratio asymptotic just proved giveFor $a>2$ the series of these probabilities converges, since its terms are eventually bounded by a constant times the convergent p-series $n^{-a/2}$. For $0<a<2$ it diverges: choose $r$ strictly between $a/2$ and one, and use $(\log n{)}^{-1/2}\ge n^{-(r-a/2)}$ for sufficiently large $n$, reducing to the divergent p-series $n^{-r}$. At $a=2$ it still diverges, as the integral test givesFor $a>2$, the first Borel-Cantelli lemma makes $A_n(a)$ occur only finitely often. For $0<a\le 2$, independence of the random variables makes the events $A_n(a)$ independent, so the second Borel-Cantelli lemma makes them occur infinitely often almost surely. Apply these conclusions simultaneously at $a=2+1/k$ for positive integers $k$, and at $a=2$. On the resulting probability-one event, the limit superior is at most every $2+1/k$ and at least two. Therefore the Gaussian sample limsup at logarithmic scale is

Let $r=|\sin x|\in [0,1]$. For $r<1$, the absolute summand $r^n/\sqrt{n}$ is at most $r^n$, so a geometric series comparison proves absolute convergence. For $r=1$, the series is $\sum (-1{)}^n/\sqrt{n}$: its positive term sizes decrease to zero, so the alternating series test proves convergence. Its absolute series is a divergent p-series with exponent $1/2$. HenceIt converges for every real $x$ and never diverges.

For a bounded real function on $[a,\infty )$, convergence of its improper integral means that its restriction to every $[a,b]$, $b>a$, is Riemann integrable and that the finite limitexists. Boundedness by itself does not guarantee either local Riemann integrability or existence of this limit.

A monotone function on $[a,b]$ is bounded between its endpoint values. Divide the interval into $m$ equal pieces with endpoints $x_j=a+j(b-a)/m$. If $f$ is nondecreasing, its upper and lower Darboux sums differ byFor nonincreasing $f$, the same calculation with reversed endpoint values gives $(b-a)(f(a)-f(b))/m$. Thus in either case $U-L=(b-a)|f(b)-f(a)|/m\to 0$. For every $\epsilon >0$ one can choose $m$ to make this gap smaller than $\epsilon$; the Riemann integrability criterion proves that $f$ is Riemann integrable.

Now let $f$ be decreasing on $[1,\infty )$ with limit zero. For each $x$, $f(x)\ge {\lim }_{y\to \infty }f(y)=0$. On $[n,n+1]$ its monotonicity givesSumming these bounds, with $S_N={\sum }_{n=1}^{N}f(n)$, yieldsBoth the partial sums $S_N$ and the truncated integrals are nondecreasing because $f\ge 0$. The inequalities show that one family is bounded above if and only if the other is. The bounded monotone sequence theorem then gives convergence of the sums exactly when the integer-endpoint integrals converge. For real $b$ between $N$ and $N+1$, the integral to $b$ lies between the integrals to these integer endpoints; the squeeze theorem supplies the same limit. If either family is unbounded, both tend to $+\infty$. This proves the integral test for convergence in the required form.

For $\alpha <0$, use $f(x)=x^{\alpha }$, which is decreasing and tends to zero. Its truncated improper integral isThis has a finite limit exactly when $\alpha <-1$. If $\alpha \ge 0$, the terms $n^{\alpha }$ do not tend to zero, so the term test for divergence applies. Thus the full p-series criterion isFor every $\alpha \ge -1$ its nonnegative partial sums tend to $+\infty$.

A power series has radius of convergence $R$ when it converges absolutely for every $|z|<R$ and diverges for every $|z|>R$. Its behaviour on $|z|=R$ is not specified by this definition. When $R=0$, it converges at the centre and diverges at every nonzero point; an infinite radius of convergence means absolute convergence at every finite $z$.

Suppose, towards a contradiction, that $|a_n{z}^n|\le M$ for all $n$ at some $|z|>R$. Choose a complex number $w$ with $R<|w|<|z|$. ThenThe comparison test for series with the convergent geometric series $\sum M{q}^n$ yields absolute convergence at $w$, contradicting the definition of $R$. Thus terms of a power series are unbounded outside its convergence disc:For the given power series, the ratio test gives, when $z\ne 0$,Hence it converges absolutely for $|z|<1$ and its terms fail to tend to zero for $|z|>1$ (the ratios are eventually larger than some number greater than $1$). Therefore $\boxed{R=1}$. In fact, on $|z|=1$ it also converges absolutely by comparison with the convergent p-series $\sum n^{-3}$.

The alternating series test states that if $b_n\ge 0$ decreases to zero, then $\sum (-1{)}^n{b}_n$ converges. Indeed, its even and odd partial sums are monotone in opposite directions and bound each other, while their difference tends to zero. Here $b_n=1/(1+\sqrt{n})$ decreases to zero, so the series converges.

It fails absolute convergence, since for $n\ge 1$,The p-series $\sum n^{-1/2}$ diverges. This also follows by comparison with the divergent harmonic series, because $n^{-1/2}\ge n^{-1}$. The comparison test for series therefore shows that $\sum b_n$ diverges. Thus

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.

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.

A real sequence $(x_n)$ has limit of a sequence $x$ whenA series ${\sum }_{n\ge 1}{x}_n$ is a convergent series with sum $s$ when its partial sums $S_N={\sum }_{n=1}^N{x}_n$ have limit of a sequence $s$.

For the first comparison test for series, either of the two allowed bounds implies $0<x_n\le a_n+b_n$. HenceThe partial sums form a monotone bounded sequence, so they converge. The series $\sum x_n$ is convergent.

For the exponent $2$, use $1/n^2\le 1/[n(n-1)]=1/(n-1)-1/n$ for $n\ge 2$. The resulting telescoping series bounds every partial sum by $2$, proving that the series converges by the monotone bounded sequence property. To prove divergence of the harmonic series, group the terms with $2^j<n\le 2^{j+1}$: their sum is at least $2^j/2^{j+1}=1/2$. Thus its partial sums are unbounded. If $\alpha \le 1$, then $n^{-\alpha }\ge n^{-1}$, so the comparison test for series proves

For the weighted square assumption, the finite-dimensional Cauchy-Schwarz inequality gives, for every $N$,Again the positive partial sums form a monotone bounded sequence. This is the weighted square roots of a summable sequence argument, with $a_n=n^2{x}_n^2$. The weighted square condition implies $\sum x_n<\infty$.

The converse is false. Take $x_n=n^{-3/2}$. Each block $2^j\le n<2^{j+1}$ contributes at most $2^j{2}^{-3j/2}=2^{-j/2}$, and the resulting geometric series converges. But $n^2{x}_n^2=1/n$, whose harmonic series diverges. This proves the counterexample without assuming the full P-series criterion.

If $s\le 0$, the terms $n^{-s}$ do not tend to zero, so the series diverges by the term test for divergence. If $s>0$, the terms form a nonnegative decreasing sequence, and the Cauchy condensation test gives the condensed seriesThis geometric series converges exactly when $1-s<0$. Hence the p-series criterion is

The Picard criterion with the input/output convention above statesNecessity follows from $⟨f,v_j⟩={\sigma }_j⟨u,u_j⟩$ and Bessel inequality. Conversely the summability constructs $u={\sum }_j{\sigma }_j^{-1}⟨f,v_j⟩u_j\in U$; the closure condition ensures that its image is all of $f$. The closure condition must not be omitted: a component orthogonal to the range cannot be reconstructed.

For the diagonal example, $f_j=j^{-p}$ is itself in ${\ell }^2$ only for $p>1/2$. Its preimage would be $u_j=j^{1-p}$, so the range criterion is the convergence of ${\sum }_j{j}^{2-2p}$. The P-series givesAt $p=3/2$ the reconstruction norm diverges harmonically. For $1/2<p\le 3/2$ the data are legitimate but have no preimage; for $0<p\le 1/2$ they are not even in the specified data space.