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If $\sum a_n$ converges and $b_n$ is bounded with $\sum |b_{n+1}-b_n|<\infty$, then $\sum a_n{b}_n$ converges. For $T_j={\sum }_{k=m}^j{a}_k$, summation by parts givesThe first factor tends uniformly to zero by the Cauchy sequence criterion, and the second has a uniform bound. A bounded monotone sequence has finite total variation, so monotone multipliers are included. Eventual monotonicity also suffices, since finitely many initial terms do not affect convergence. Absolute convergence of $\sum a_n$ is unnecessary.

For smooth non-increasing $V$ and an ordered subsolution and supersolution, begin with $u_0={\phi }^{-}$ and solve successive Poisson equations with forcing $V(u_k)$ and the fixed boundary data. The weak maximum principle for elliptic operators applied to the negatives of consecutive differences givesThe global Schauder estimate and the Hölder interpolation inequality bound the iterates uniformly in $C^{2,\alpha }$. Their bounded monotone sequence converges, and compact embedding of Hölder spaces permits passing to the equation. The uniform Hölder seminorm bounds retain $C^{2,\alpha }$ regularity of the limit.

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.

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.

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.

Because $(a_n)$ is a monotone sequence, every one of its first $n$ terms is at most $a_n$, soOn the other hand, every term from $a_n$ through $a_{2n}$ is at least $a_n$, and henceBoth the lower bound and upper bound tend to $x$ because the Cesaro means converge to $x$. The squeeze theorem gives

Use the interior ordered subsolution and supersolution correction from part (a). Recursively solve the Poisson equation $\Delta u_{k+1}=V(u_k)$ with boundary value $\psi$, starting from $u_0={\phi }^{-}$. Part (a) gives $u_0\le u_1\le {\phi }^{+}$. If $u_{k-1}\le u_k\le {\phi }^{+}$, thenThe first difference has zero boundary values, the second nonnegative values. The weak maximum principle for elliptic operators applied to their negatives gives $u_k\le u_{k+1}\le {\phi }^{+}$. Induction proves the bounded monotone sequenceWithout interior ordering, the counterexample in part (a) disproves the claim.

With the interior ordering correction in part (a), iteration of part (c) givesThe bounded monotone sequence has a pointwise limit $u$. By compact embedding of Hölder spaces, a subsequence converges in $C^{2,\beta }$ for $0<\beta <\alpha$, necessarily to this same limit. All convergent subsequences have that limit, so compactness gives convergence of the full sequence in $C^{2,\beta }$. Passing to the limit in $\Delta u_k=V(u_{k-1})$ gives the equation and boundary values. The uniform Hölder seminorm bounds on second derivatives pass to the limit, so $u\in C^{2,\alpha }(\overline{\Omega })$, and the barriers remain ordered:This proves the intended monotone iteration for a semilinear elliptic equation. With only the boundary ordering printed in the PDF, the counterexample in part (a) shows the requested trapped solution need not exist.

The Fibonacci recurrence givesThe map $x\mapsto 1+1/x$ is strictly decreasing for $x>0$. Since $a_3=2\ge a_1=1$, applying this decreasing map reverses each inequality and proves by mathematical induction thatTaking even $n$ shows $a_{2n+2}\le a_{2n}$, so $(a_{2n})$ is a decreasing sequence.