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Knowing the radius of convergence does not settle convergence on its boundary. The three power seriesall have radius one, but behave differently on $|z|=1$.

For the geometric series, boundary terms all have complex modulus one, violating the term test for divergence. For the second series, absolute values give the divergent harmonic series at every boundary point. At $z=1$ the series itself is harmonic and diverges. At any other point on the unit circle, its geometric partial sums obeyThe Dirichlet test with the decreasing sequence $1/n\to 0$ therefore gives conditional convergence. In particular, $z=-1$ gives the negative of the alternating harmonic series, with sum $-\ln 2$. For the third series, the convergent p-series ${\sum }_{n\ge 1}1/n^2$ bounds the absolute values at every boundary point, giving absolute convergence by the comparison test for series.

At $z=-1$, the rational function $1/(1-z)$ is well defined and equals $1/2$, yet its geometric series diverges there. A formula for a function and a particular series representation can have different domains. Likewise, applying termwise differentiation of a power series preserves the radius but need not preserve the boundary convergence.

The fundamental strip of a Mellin transform is the interior of the set of real parts $\sigma$ for whichViewed in the complex plane, it is a vertical strip $\alpha <\mathrm{Re}s<\beta$; either bound may be infinite, and the interior may be empty. Both endpoints of the improper integral must be checked. Convergence on a boundary line, whether absolute convergence or conditional convergence, requires a separate test. The Mellin transform is holomorphic throughout a nonempty fundamental strip.^[ref]

A practical sufficient test uses Big O notation: if $f(x)=O(x^p)$ near zero and $f(x)=O(x^{-q})$ near infinity, with real $p,q$, then comparison with elementary improper integrals givesThus $-p<\sigma <q$ guarantees absolute convergence. These upper bounds on $|f|$ need not locate the maximal Mellin fundamental strip; faster actual decay can widen it. The strip property itself follows by bounding the weight at any intermediate $\sigma$ by the sum of two weights already known to be integrable. On a closed narrower strip, neighboring integrable weights also dominate the natural logarithm factors introduced by differentiation under the integral sign, making the transform a holomorphic function.

An exact two-endpoint example isHere both bounds are sharp: at $\sigma =-2$ the zero-end absolute convergence test contains ${\int }_0^{1}dx/x$, and at $\sigma =3$ the infinity-end test contains ${\int }_1^{\infty }dx/x$. The displayed rational function continues meromorphically outside the Mellin fundamental strip and has simple poles at $-2$ and $3$. Its other finite continued values outside the strip are values of a meromorphic continuation, not of a convergent defining integral. More general endpoint information leads to meromorphic continuation of a Mellin transform from an asymptotic expansion.

Integrating the geometric series for $1/(1+x)$ from zero to $x$ gives the displayed power series for $|x|<1$. It has radius of convergence one. At $x=1$ the alternating series test gives conditional convergence to $\ln 2$; at $x=-1$ the negative harmonic series diverges. Replacing $x$ by $-w$ gives $\ln (1-w)=-{\sum }_{n\ge 1}{w}^n/n$. An expression formed by substituting a nonlinear function for $w$ need not have the same convergence range after it is re-expanded as a Taylor series in the new variable.

Set $c_n=4^{-n}\left(\genfrac{}{}{0ex}{}{2n}{n}\right)$, with $c_0=1$. Cancellation of factorials givesConsequently $c_n$ decreases to zero: $\log c_n\le -\frac{1}{2}{\sum }_{j=1}^{n}1/j\le -\frac{1}{2}\log (n+1)$. The inequality uses $\log (1-x)\le -x$ and comparison with ${\int }_1^{n+1}dt/t$. Conversely $c_n\ge 1/(n+1)$ by induction, since its ratio is at least $(n+1)/(n+2)$. Thus $\sum c_n$ diverges while $\sum (-1{)}^n{c}_n$ has conditional convergence by the alternating series test. These elementary bounds require no asymptotic formula for factorials.

The terms at $n=2j$ and $n=2j+1$ are $\ln [(2j+1)/(2j)]$ and its negative. Thus the partial sums ending at an odd index are zero, while those ending at $2m$ equal $\ln [(2m+1)/(2m)]\to 0$. Considering both subsequences proves convergence of the full sequence of partial sums, not merely of grouped terms.

The paired sum of absolute values is $2\ln [1+1/(2j)]$. Since $\ln (1+u)\ge u/(1+u)$ for $u\ge 0$, these pairs dominate $2/(2j+1)$, whose series diverges by comparison with the harmonic series. The original series therefore has conditional convergence with the displayed sum.

The positive terms $b_n=1/\log (n+1)$ decrease to zero. The alternating series test just proved therefore gives convergence of the signed series. It is conditional convergence: $\log (n+1)\le n$, so $b_n\ge 1/n$, and the absolute-value series dominates the divergent harmonic series.

The comparison test for series used here says that if $0\le c_n\le d_n$, convergence of $\sum d_n$ implies convergence of $\sum c_n$, while divergence of $\sum c_n$ implies divergence of $\sum d_n$. This follows by comparing increasing partial sums. For completeness, the harmonic series diverges because every block from $2^k$ to $2^{k+1}-1$ has sum at least $1/2$.

The alternating series test says that if $a_n\ge 0$ decreases to $0$, thenconverges. Taking $a_n=n^{-1/2}$ proves convergence of the given series.

It is not absolute convergence, because the series of absolute values is the p-serieswhich diverges since $p=1/2\le 1$. Thus the original series has conditional convergence.

For an explicit divergent rearrangement of a series, take unused positive terms, which are the even-indexed terms, until the partial sum exceeds $1$, then take the first unused negative term. Next take positive terms until the sum exceeds $2$, then the next unused negative term, and continue. Both the positive and negative subseries have infinite total magnitude, so this procedure uses every term. The negative term inserted at stage $m$ tends to zero, while the preceding partial sum exceeds $m$; consequently these rearranged partial sums tend to $+\infty$. This is a divergent series with exactly the prescribed terms.

For $p>0$, the function $x\mapsto \sin x/x^p$ is Lebesgue integrable on $(0,\infty )$ exactly when $1<p<2$. Near zero, $\sin x/x\to 1$, so absolute integrability is equivalent to that of $x^{1-p}$, which requires $p<2$. At infinity, $p>1$ gives the upper bound $x^{-p}$. For $0<p\le 1$, on each interval $[k\pi ,(k+1)\pi ]$,and the corresponding P-series diverges.

The signed improper integral nevertheless converges for every $0<p<2$: near zero the same local comparison applies, and at infinity the Dirichlet test for improper integrals uses the bounded antiderivative of sine and the decreasing factor $x^{-p}\to 0$. Thus $0<p\le 1$ gives conditional convergence, while $1<p<2$ gives absolute integrability. In particular, substituting $t=1/x$ turns ${\int }_0^1|x^{-1}\sin (1/x)|dx$ into the divergent tail ${\int }_1^{\infty }|\sin t|/tdt$.