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For $x\ge 0$, the function $h(x)=e^x-1-x$ has $h(0)=0$ and $h^{\prime }(x)=e^x-1\ge 0$, because the exponential function is increasing and $e^0=1$. Therefore $\boxed{e^x\ge 1+x}$.

Write $S_n={\sum }_{j=1}^n{a}_j$ and $P_n={\prod }_{j=1}^n(1+a_j)$. Expanding the product produces $1+S_n$ and additional nonnegative terms, giving $P_n\ge 1+S_n\ge S_n$. Applying the exponential inequality to every factor givesBoth sequences are increasing, since the terms $a_j$ are positive. If $S_n$ converges to a finite value $S$, then $P_n\le e^S$, so the convergence theorem for monotone bounded sequences makes $P_n$ converge. Conversely, if $P_n$ has a finite limit, then $S_n\le P_n$ is bounded and increasing, hence converges by the same theorem. Thus the positive series and the product both converge to finite limits, or both diverge to $+\infty$. This proves the positive sum-product convergence criterion. The word limit here means a finite real limit; allowing extended infinite limits would make the conclusion vacuous for these monotone sequences.

Let $s_N={\sum }_{n=1}^N(-1{)}^{n-1}{b}_n$. Pairing consecutive terms shows $s_{2m}\ge 0$ andThe even partial sums increase and the odd partial sums decrease. They satisfy $s_{2m}\le s_{2m+1}\le b_1$, whileBoth are monotone bounded sequences and hence have limits, and the displayed difference makes their limits equal. Every partial sum belongs to one subsequence, so the alternating series converges. This is a proof of the alternating series test; the monotonicity and zero-limit hypotheses play separate roles.

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.

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 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, $(n\log n)x_n=1$.

Let $(u_n)$ be increasing and bounded above, and let $L=\sup \{u_n:n\in \mathbb{N}\}$. For every $\epsilon >0$, the definition of the supremum supplies $N$ with $L-\epsilon <u_N\le L$. Monotonicity gives $L-\epsilon <u_n\le L$ for every $n\ge N$, proving $u_n\to L$. This is the monotone bounded sequence theorem.

For the recurrence, $f(x_n)>0$ makes $(x_n)$ increasing. If it were bounded above, it would converge to some $L$. Since $x_n\le L$ and $f$ is decreasing, $f(x_n)\ge f(L)>0$, whencea contradiction. Therefore $x_n\to \infty$.

Part (i) also shows that the odd subsequence is increasing and the even subsequence is decreasing. Each odd term is below the adjacent even term, so both are bounded monotone sequences and therefore have limits. Part (ii) says their difference tends to zero, so those limits coincide. Interlacing the two subsequences proves that $(a_n)$ has that common limit.

Every convergent sequence is bounded. Conversely, let $(x_n)$ be increasing and bounded above, and set $L=\sup \{x_n:n\ge 1\}$. For any $\epsilon >0$, $L-\epsilon$ is not an upper bound, so some $x_N>L-\epsilon$. Monotonicity then givesfor all $n\ge N$. Hence $|x_n-L|<\epsilon$, proving the monotone bounded sequence criterion.

For $a_j\ge 0$, the finite product satisfiesThe lower inequality follows by product expansion; the upper one uses $1+t\le e^t$. Since both the partial sums and partial products are increasing, one converges to a finite limit exactly when the other does. The monotone bounded sequence theorem proves both implications. The product cannot tend to zero in this nonnegative setting.