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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

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 Cesàro means converge to $x$. The squeeze theorem gives

Componentwise,soRepeated application gives $(A^k{)}^T=(A^T{)}^k$. Transposing the matrix exponential term by term therefore yieldsMultiplication of the two power series givesHenceIf $e^{tA}$ is an orthogonal matrix for every real $t$, thenIts coefficient of $t$ must vanish, sothat is, $A$ is a skew-symmetric matrix.

Let $L$ be the supremum of all lower Darboux sums and $U$ the infimum of all upper Darboux sums. The assumed comparison of arbitrary lower and upper sums gives $L\le U$. For the partition supplied for a given $\epsilon$,Since this holds for every positive $\epsilon$, $U=L$. Thus the Riemann integrability criterion gives

If $f$ is continuous on $[a,b]$, the Heine-Cantor theorem makes it uniformly continuous. Given $\epsilon >0$, choose $\delta$ so that intervals of length below $\delta$ have oscillation below $\epsilon /(b-a)$. A partition of mesh below $\delta$ then satisfiesproving that Continuous functions are Riemann integrable.

For the function defined using $g$, choose $K$ with $|f|\le K$. Given $\epsilon >0$, choose $\delta >0$ so small that $2K\delta <\epsilon /2$. On $[\delta ,1]$, the function $g$ is continuous and therefore has a partition whose upper-minus-lower sum is below $\epsilon /2$. Adding the interval $[0,\delta ]$ contributes at most $2K\delta$, regardless of the value $f(0)=\lambda$. Hence the full Darboux-sum difference is below $\epsilon$, and

Finally, for any partition $a=x_0<\cdots <x_n=b$, the mean value theorem supplies ${\xi }_i\in (x_{i-1},x_i)$ such thatSumming telescopes toa Riemann sum for the Riemann-integrable function $f^{\prime }$. Equivalently, every lower derivative sum is at most this telescoping value and every upper derivative sum is at least it. Taking the common upper and lower Darboux integrals proves the fundamental theorem of calculus conclusion