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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.

Rationalize the numerator of the summand:For every $n\ge 1$ we have $\sqrt{1+1/n}\le \sqrt{2}$, and thereforeThe harmonic series diverges, so the comparison test for series makes the given positive-term series diverge as well:For completeness, divergence of the harmonic series follows by grouping the terms with $2^k<n\le 2^{k+1}$: each such block contributes at least $2^k/2^{k+1}=1/2$. The same rationalization also shows that $n$ times the summand tends to $1/2$. The summands tend to zero, but their harmonic-size tail prevents convergence.

Every counted point has integer coordinates and lies in the square $[-n,n{]}^2$, while the origin is always counted. HenceNo precise circle-area estimate is needed; this elementary bound controls the growth of the power series coefficients.

The comparison test for series states that if $0\le u_n\le v_n$, convergence of $\sum v_n$ implies convergence of $\sum u_n$. Equivalently, divergence of the smaller nonnegative series forces divergence of the larger one.

For $|z|=r<1$,For $0<r<1$, the ratio of successive terms on the right tends to $r<1$, so the ratio test and the comparison test for series show absolute convergence. The case $r=0$ is immediate.

If $|z|\ge 1$, then $|a_n{z}^n|\ge 1$, so the terms do not tend to zero. The term test for divergence therefore rules out convergence, including every point on the boundary $|z|=1$. Thus the radius of convergence isThe series converges exactly in the open unit disc. This is an example of polynomial coefficient growth with unit power-series radius: the lattice-count bound is polynomial rather than exponential.

For a nonempty set $S\subseteq \mathbb{R}$, an upper bound in a partially ordered set is a real number $u$ such that $s\le u$ for every $s\in S$. Its supremum, or least upper bound, is an upper bound $L$ with $L\le u$ for every upper bound $u$. The least-upper-bound property asserts that every nonempty set of real numbers bounded above has a real supremum.

Let $(x_n)$ be a bounded, increasing sequence; here increasing allows $x_{n+1}=x_n$. Apply the least-upper-bound property to its set of terms and put $L=\sup \{x_n:n\ge 1\}$. Given $\epsilon >0$, the number $L-\epsilon$ cannot be an upper bound, so some $x_N$ satisfies $x_N>L-\epsilon$. The monotone sequence property then givesHence $|x_n-L|<\epsilon$ for all $n\ge N$. This proves the monotone bounded sequence result, with limit $\boxed{x_n⟶L}$.

For the two series, define their partial sums $A_N={\sum }_{n=1}^N{a}_n$ and $B_N={\sum }_{n=1}^N{b}_n$. Positivity makes both partial-sum sequences increasing, and termwise comparison gives $0<B_N\le A_N$. Since $(A_N)$ is a convergent sequence, it is bounded above, and therefore so is $(B_N)$. Applying the monotone bounded sequence result already proved shows that $(B_N)$ converges. By the definition of a convergent series, ${\sum }_{n=1}^{\infty }{b}_n$ converges. This is the comparison test for series for positive terms.

The arithmetic-geometric mean inequality gives $a_k+1/a_k\ge 2$, so $u_n\ge 2n\to +\infty$. Hence $(u_n)$ has no finite limit.

For the example, take $\boxed{a_k=1+2^{-k}}$. Every term is positive and different from $1$, andThe bounding geometric series converges, so the comparison test for series proves that the partial sums $v_n$ converge.

In general, if $v_n$ converges, the necessary condition for a convergent series gives $d_k=a_k-1/a_k=v_k-v_{k-1}\to 0$. Since $a_k>0$,Apply the Cesàro theorem for convergent sequences to these summands:Equivalently, $a_k=(d_k+\sqrt{d_k^2+4})/2\to 1$. The series of the differences may converge while the series of the positive sums still diverges linearly.

The comparison test for series states that if $0\le a_n\le b_n$ eventually and $\sum b_n$ converges, then $\sum a_n$ converges. The partial sums of $\sum a_n$ are increasing and bounded above by a constant plus the bounded partial sums of $\sum b_n$, so the monotone bounded sequence theorem applies.

If $\sum x_n$ converges, then $x_n^2\le x_n$ proves convergence of $\sum x_n^2$. Also $x_n\to 0$, so eventually $x_n\le 1/2$ and $x_n/(1-x_n)\le 2x_n$, proving convergence of $\sum x_n/(1-x_n)$. The converse for the square series is false: $x_n=1/(n+1)$ has convergent square series but divergent original series. The converse for the fraction series is true because $x_n\le x_n/(1-x_n)$.

If $(x_n)$ is positive and decreasing and $\sum x_n$ converges, put $m=\lfloor n/2\rfloor$. Thenso $n{x}_n\to 0$. The converse fails for $x_n=1/(n\log n)$ for $n\ge 2$: it is decreasing and $n{x}_n\to 0$, but the integral test for convergence shows that its series diverges.

Even convergence of $\sum x_n$ need not imply $(n\log n)x_n\to 0$. Choose integers $N_j=\lceil e^{2^j}\rceil$ and set $x_n=(N_j\log N_j{)}^{-1}$ for $N_{j-1}<n\le N_j$. This is decreasing, and the $j$th block contributes at most $1/\log N_j\le 2^{-j}$, so the series converges. At $n=N_j$, however, $(n\log n)x_n=1$.

For positive $a_n,b_n$,The comparison test for series therefore gives

The arithmetic-geometric mean inequality givesThus convergence of ${\sum }_n{a}_n$ implies convergence of ${\sum }_n\sqrt{a_n{a}_{n+1}}$ by the comparison test for series.

The converse is false. DefineThe even terms make ${\sum }_n{a}_n$ diverge, whileso the neighboring geometric-mean series converges.

If $z_0=0$, there is no $z$ with $|z|<|z_0|$, so the claim is vacuous. Suppose $z_0\ne 0$. Convergence of $\sum a_n{z}_0^n$ implies $a_n{z}_0^n\to 0$, because terms of a convergent series tend to zero. In particular this is a bounded sequence, so $|a_n{z}_0^n|\le C$ for some finite $C$ and every $n$.

For $|z|<|z_0|$, put $q=|z|/|z_0|<1$. ThenThe geometric series and comparison test for series proveThus convergence at one point gives absolute convergence at every point strictly closer to the centre.

The comparison test for series states that, for $0\le u_n\le v_n$ eventually, convergence of $\sum v_n$ implies convergence of $\sum u_n$. Equivalently, divergence of $\sum u_n$ implies divergence of $\sum v_n$.

For $|z|<1$, the coefficient bound givesby the geometric series formula and the comparison test for series. The single term $a_0$ does not affect convergence. For $|z|\ge 1$,so the terms do not tend to zero. The fact that terms of a convergent series tend to zero excludes convergence there. Thus

Use the ceiling function to choose positive integer coefficientsBecause $0<\alpha <1$, we have ${\alpha }^{-n}\ge 1$ andFor $|z|<\alpha$, the comparison test for series with the geometric series givesFor $|z|\ge \alpha$, we instead have $|a_n{z}^n|\ge (|z|/\alpha {)}^n\ge 1$, so the terms do not tend to zero. Thus the power series has radius of convergence exactly $\alpha$.

All three examples below have positive integer coefficients, so they work under either convention for whether $0\in \mathbb{N}$.

For $R=1$, take $\boxed{a_n=1}$. This is the geometric series: it converges for $|z|<1$, and its terms fail to tend to zero for $|z|\ge 1$.

For $R=0$, take $\boxed{a_n=n!}$, using the factorial and $0!=1$. For every fixed $z\ne 0$, the absolute values of successive terms have ratio $(n+1)|z|$. This eventually exceeds two, so the terms grow without bound and the series cannot converge. At $z=0$ it is finite, giving radius of convergence zero.

For $R=1/\sqrt{2}$, takeusing the floor function. The coefficients satisfyThe upper bound and the comparison test for series give absolute convergence when $\sqrt{2}|z|<1$. The lower bound prevents the terms tending to zero when $\sqrt{2}|z|\ge 1$. Hence the radius of convergence is exactly $1/\sqrt{2}$.

For every real $\alpha$, $|\cos (\alpha n)/n^2|\le 1/n^2$. The bounding P-series with exponent two converges. The comparison test for series therefore givesOscillations of cosine are unnecessary for convergence here.

Rationalize the difference to reveal its small size instead of comparing its two large summands separately:For any fixed real $\alpha$, the right-hand series is a constant multiple of a convergent P-series. The comparison test for series provesThe case $\alpha =0$ is the identically zero series. For nonzero $\alpha$, its summand is asymptotic to ${\alpha }^4/(2n^2)$, explaining the same convergence behavior as the inverse-square series.

The comparison test for series states that if $0\le a_n\le b_n$ for all sufficiently large $n$, then convergence of $\sum b_n$ implies convergence of $\sum a_n$; conversely, divergence of $\sum a_n$ implies divergence of $\sum b_n$. A summable bound on $|a_n|$ proves absolute convergence of $\sum a_n$.

For the ratio test, suppose the terms are eventually nonzero and $L={\lim }_{n\to \infty }|a_{n+1}/a_n|$ exists. If $L<1$, the series converges absolutely. If $L>1$, including $L=\infty$, the terms fail to tend to zero and the series diverges. If $L=1$, the test is inconclusive. An eventually zero series converges separately. Convergence means that the sequence of partial sums has a finite limit; the terms tending to zero is only a necessary condition for a convergent series.

The comparison test for series says that $0\le v_n\le C{w}_n$ eventually and convergence of $\sum w_n$ imply convergence of $\sum v_n$. Conversely, an eventual positive lower comparison with a divergent positive series proves divergence. The ratio test uses $L={\lim }_{n\to \infty }{v}_{n+1}/v_n$: $L<1$ proves convergence, $L>1$ proves divergence, and $L=1$ is inconclusive. Removing finitely many initial zero terms does not affect these conclusions.

The comparison test for series states that if $0\le u_n\le v_n$ for all sufficiently large $n$, convergence of $\sum v_n$ implies convergence of $\sum u_n$. Equivalently, divergence of the smaller nonnegative series forces divergence of the larger one. Finitely many initial terms do not affect either convergence conclusion.

Use the bound $|\sin n|\le 1$:The comparison p-series has exponent $2>1$ and converges. Hence, by the comparison test for series, the original series is absolutely convergent. No alternating-sign or monotonicity assumption about $\sin n$ is needed.

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.