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The limit comparison test says that if $u_n,v_n>0$ and $u_n/v_n\to L$ with $0<L<\infty$, then their series either both converge or both diverge. This follows because, eventually, $(L/2)v_n\le u_n\le (3L/2)v_n$, and the comparison test for series applies in both directions.

Here comparison with the harmonic series givesFor completeness, the natural logarithm grows slowly enough because for $x\ge 1$,so $(\log n{)}^2/n\le 16/\sqrt{n}\to 0$. The harmonic series diverges: every block from $n=2^k$ to $2^{k+1}-1$ has $2^k$ terms at least $2^{-k-1}$ and therefore contributes at least $1/2$. Hence

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 $s=\pm 1$, the identity $s/(2+s)=-1/3+(2/3)s$ givesThe alternating series test applies to $\sum (-1{)}^n/n$ because $1/n$ decreases to zero; its partial sums consequently have a finite limit. The harmonic series $\sum 1/n$ diverges to $+\infty$. Applying the displayed identity to each finite partial sum yieldsTherefore $\boxed{\text{the series diverges to }-\infty }$. Alternating signs alone do not satisfy the alternating series test: the magnitudes jump from $1/(3n)$ at even $n$ to $1/(n+1)$ at the next odd index.

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.

The summand is positive, so it suffices to show that its partial sums are unbounded. On the block $2^k\le n<2^{k+1}$, for $k\ge 1$,There are $2^k$ terms in that block, so its sum is at least $1/[2(k+1)\log 2]$. Adding blocks gives a constant multiple of the divergent harmonic series. HenceEquivalently the integral test for convergence gives ${\int }_2^{R}dx/(x\log x)=\log \log R-\log \log 2\to \infty$. The block proof supplies the divergence without assuming the integral test.

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.

For a planar simple symmetric random walk write its coordinates as $(S_n,T_n)$ and rotate to $(S_n+T_n,S_n-T_n)$. Each step now has four equally likely values $(1,1),(1,-1),(-1,1),(-1,-1)$. Thus the two rotated coordinates are independent one-dimensional simple symmetric random walks.

A return is impossible at odd times, and at time $2n$ its probability isFor $n\ge 2$, the supplied estimate gives $1/(4n)<p_{2n}(0,0)<1/n$. At $n=1$ the lower bound is actually equality, $p_2(0,0)=1/4$; this harmless strict-endpoint typo does not affect the argument. The harmonic series diverges, so the criterion proved above shows that the origin is recurrent, in the terminology of a recurrent Markov chain. Translation invariance gives the same conclusion for every state. The Markov chain is irreducible, so from any starting point it hits any fixed state almost surely and revisits it infinitely often. One can see the hitting assertion directly: repeated returns to the starting point give repeated positive-probability opportunities to follow a finite path to the target.

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.

Convergence at $z_0$ implies that the terms $a_n{z}_0^n$ form a bounded sequence: $|a_n{z}_0^n|\le M$ for some $M$. If $|z|<|z_0|$ and $r=|z/z_0|<1$, thenComparison with the convergent geometric series $\sum M{r}^n$ proves absolute convergence. The case $z_0=0$ is vacuous.

The radius of convergence is the number $R\in [0,\infty ]$ such that a power series converges absolutely for $|z|<R$ and diverges for $|z|>R$, with boundary behavior considered separately.

TakeThen $\sum v^n/n$ is the convergent alternating harmonic series, whereas $\sum w^n/n$ is the divergent harmonic series. Thus $\sum z^n/n$ converges at a point of modulus one and diverges at another point of modulus one. It follows that

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 Cesàro theorem for convergent sequences. Its hypothesis concerns a convergent sequence; the corresponding series need not be a convergent series.

For $x\ge 2$, expand the finite Euler product as a product of geometric series:Every integer $1\le n\le x$ has all its prime factors at most $x$, soThe harmonic series diverges, and therefore

Suppose instead that ${\sum }_{p}1/p$ converged. For $0<t\le 1/2$,Since ${\sum }_{p}1/p^2$ converges, this would makeconverge to a finite limit, contradicting the divergence of $P(x)$. Hence the Euler proof that the sum of reciprocals of primes diverges gives

At the cusp of $f_0(t)=|t|$, positivity and evenness of the Fejér kernel giveSince $\sin (t/2)\le t/2$,Summing the supplied lower bound over $I_k$, $1\le k\le n^{\prime }=\lfloor n/2\rfloor$, yieldsThe harmonic series satisfies ${\sum }_{k=1}^{n^{\prime }}{k}^{-1}\ge c^{\prime }\log n$, so

The modulus of continuity of the periodic function $f_0$ obeys $\omega (f_0,1/n)=1/n$. If a universal Jackson-type estimateheld for all continuous periodic $f$, it would give $O(1/n)$ for $f_0$, contradicting the lower bound. Thus

For $k\ge 1$, setThen $t_k\downarrow 0$ and $f(t_k)=(-1{)}^k{t}_k$. Consecutive values have opposite signs, soFinite partitions containing $t_N,t_{N-1},\dots ,t_1$ therefore have variation at leastwhich diverges with $N$ by comparison with the harmonic series. Part (b) now gives $\Vert f\Vert =\infty$.

For a primitive nonprincipal Dirichlet character modulo $q$, finite Fourier inversion theorem and the primitive Gauss magnitude give this bound uniformly in the interval. Bound the exponential sums by $C\Vert a/q{\Vert }^{-1}$ and sum the resulting harmonic series. The general Gauss identity follows from the prime-power identity by the Chinese remainder theorem.

For $u,v\ge 0$ with $u+v>0$, multiplication by the sum of the square roots givesThis identity exposes small differences hidden by cancellation. For example, $\sqrt{n^2+n}-n=1/(\sqrt{1+1/n}+1)\to 1/2$. It also converts normalized square-root increments into terms comparable with the harmonic series.

A series adds the terms of a sequence in their specified order. Its infinite-sum notation is defined through finite partial sums:when the displayed limit of a sequence exists as a finite number. Otherwise the series is not a convergent series in the ordinary sense. Writing an infinite list of additions does not by itself assign it a finite sum. Cesàro summation is a different prescription, with its own definition.

Two positive series make the distinction concrete. For $a_n=1/n$, the individual terms tend to zero but the harmonic series has unbounded partial sums. For $a_n=1/[n(n+1)]$, partial fraction decomposition reveals a telescoping series whose partial sums tend to one. The useful question is how much all the remaining terms can contribute together, rather than only how small the next term looks.

Choose one random variable $U$ with the continuous uniform distribution on $[0,1]$, and reuse it in every event:Each probability is $1/(n+1)$, so their harmonic series diverges, just as in summable and harmonic event probabilities. But the events are decreasing: $A_{n+1}\subseteq A_n$. For every outcome $U=u>0$, all indices with $n+1>1/u$ fail. Only finitely many events occur on that outcome. The limsup of events is exactly $\{U=0\}$, a zero-probability event. This is a concrete closed-interval version of the shrinking-interval dependence example.^[ref]

The missing assumption is visible already in two events. For $i<j$, their intersection is $A_j$, henceMoreover, the whole tail union is just $A_m$, with probability $1/(m+1)\to 0$. This calculation explains directly why divergence alone fails: overlapping events keep spending their individual probabilities on the same outcomes.

Let $N={\sum }_{n\ge 1}{1}_{A_n}$ be the total number of occurrences, a sum of indicator random variables. Although $N<\infty$ almost surely, Tonelli theorem gives $\mathbb{E}[N]={\sum }_{n\ge 1}1/(n+1)=\infty$. More explicitly, $\mathbb{P}(N\ge k)=1/(k+1)$ for $k\ge 1$: very small positive values of $U$ give very long, but finite, initial runs. Infinite expected value does not imply a positive probability of an infinite count.