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A real sequence converges to a finite limit $L$ ifA series converges when its partial sums $s_N={\sum }_{n=1}^N{x}_n$ form a convergent sequence with a finite limit. These are different conditions: a series concerns accumulated terms, whereas sequence convergence concerns the terms themselves.

If $s_N\to S$, then $s_{N-1}\to S$ as well and $x_N=s_N-s_{N-1}\to 0$. Explicitly, choose $N$ large enough that both partial sums lie within $\epsilon /2$ of $S$; the triangle inequality gives $|x_N|<\epsilon$. Hence convergence of a series forces its terms to tend to zero.

For the converse, take $x_n=1/n$. This is a null sequence, but its harmonic series diverges. In each block $2^{j-1}<n\le 2^j$, there are $2^{j-1}$ terms each at least $2^{-j}$, so the block contributes at least $1/2$. ThusA zero limit for the terms is necessary, but not sufficient, for convergent series.

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.

A sequence $(a_k)$ converges to $a\in \mathbb{R}$ when for every $\epsilon >0$ there is $N$ such that $|a_k-a|<\epsilon$ for all $k\ge N$. A series ${\sum }_{k=1}^{\infty }{a}_k$ converges to a finite $S$ when its partial sums $s_n={\sum }_{k=1}^n{a}_k$ converge to $S$.

If the series converges, then $a_n=s_n-s_{n-1}\to S-S=0$, so its terms tend to zero. The converse is false: the harmonic series has terms tending to zero but diverges.

For the arithmetic averages, suppose $a_k\to a$. Fix $\epsilon >0$ and choose $K$ so $|a_k-a|<\epsilon /2$ for $k\ge K$. The finite initial contribution $C={\sum }_{k=1}^{K-1}|a_k-a|$ is fixed, andFor sufficiently large $n$, $C/n<\epsilon /2$, provingThis is the Cesaro theorem for convergent sequences. Its hypothesis concerns a convergent sequence; the corresponding series need not be a convergent series.

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 Cesaro 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.

LetIf $x$ were rational, thenwould make $\sqrt[3]{6}=(x^3-5)/(3x)$ rational. But if $\sqrt[3]{6}=a/b$ in lowest terms, then $a^3=6b^3$, which is impossible by unique prime factorization: the exponent of $2$ on the two sides is respectively a multiple of three and one more than a multiple of three. Therefore

Use the given convergent seriesSuppose this number were $a/b\in \mathbb{Q}$. Choose $N$ large enough that $b\mid (2N+1)!$. Multiplication by $(2N+1)!$ makes both the rational number and the partial sum through $n=N$ integers. Their differencewould therefore be an integer. It is positive, whilefor sufficiently large $N$, a contradiction. Hence

A transcendental number is a complex number that is not a root of any nonzero polynomial with rational, equivalently integer, coefficients. LetIf $a\ne 0$ and $y$ were an algebraic number, then $e$ would satisfyover the algebraic extension $\mathbb{Q}(y)$. By transitivity of algebraic extensions, $e$ would be algebraic over $\mathbb{Q}$, contradicting its transcendence. If $a=0$, then $b\ne 0$ and $e=b/y$ would again be algebraic. Thus

A sequence $(a_n)$ is summable when the series ${\sum }_n{a}_n$ is convergent.