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Let $S_N={\sum }_{n=1}^N{f}_n$ and let $s$ be its uniform limit. Choose a finite $B\ge 0$ with $|g(x)|\le B$ for all $x\in E$. The partial sums of the multiplied series are $g{S}_N$, andFor any $\epsilon >0$, uniform convergence supplies $N_0$ making $|S_N(x)-s(x)|<\epsilon /(B+1)$ for all $N\ge N_0$ and $x\in E$. The displayed estimate then gives an error less than $\epsilon$. Thus

Under the stated piecewise smooth hypotheses, the Dirichlet convergence theorem for Fourier series gives the average of the two one-sided limits of the corresponding periodic extension. The even extension is continuous at both endpoints, so the Fourier cosine series converges toThe odd extension agrees with $f$ on $(0,L)$, but its left and right limits are $-f(0),f(0)$ at zero and $f(L),-f(L)$ at the join $L$. Therefore the Fourier sine series converges toIndeed every partial sum of the Fourier sine series is already zero at both endpoints. The endpoint values agree with $f$ only when the corresponding boundary values vanish.

First, the Fourier coefficients of the Fourier cosine series decay as $O(n^{-2})$, whereas those of the Fourier sine series are of order $n^{-1}$. In particular, the absolute convergence bound for the Fourier cosine series gives a uniform tail estimateThus its tail is absolutely controlled using fewer Fourier harmonics. The Fourier sine series does not have an absolutely summable sequence of Fourier coefficients.

Second, the even extension is continuous, so its Fourier cosine series converges uniformly on the closed interval and approximates the nonzero endpoint limits correctly. The odd periodic extension has jumps, producing the Gibbs phenomenon in its Fourier sine series near zero and $L$. Every Fourier sine series partial sum is zero at both endpoints, so by approaching $L$ from inside $(0,L)$ one obtainsHence the Fourier sine series approximation cannot converge uniformly even on the whole open interval, despite pointwise convergence at each interior point. The Fourier cosine series approximation has faster coefficient decay and avoids the persistent errors near the endpoints.

Set $u=x/L$ and $q=n\pi$. The required integrals follow fromConsequently the Fourier coefficients areThus the Fourier cosine series isand its sum is $e^{x/L}$ everywhere on $[0,L]$. The Fourier sine series isand its sum is $e^{x/L}$ on $(0,L)$ and zero at the two endpoints. The open and filled circles in the sketches distinguish these endpoint limits from the actual values of the sum.

Take $\boxed{p(x)=x(L-x)}$. It vanishes at both endpoints. Its odd periodic extension is continuously differentiable across the joins, whereas its even extension has corners because $p^{\prime }(0)=L$ and $p^{\prime }(L)=-L$ are nonzero.

Two integrations by parts, with $q=n\pi /L$, giveThe Fourier coefficients are thereforeThe Fourier sine series has a uniform absolute tail $O(N^{-2})$, and each of its partial sums satisfies both zero boundary values exactly. The Fourier cosine series has an absolute tail bound $O(N^{-1})$; at $x=0$ its omitted even-index Fourier coefficients have the same sign and yield an error of order $N^{-1}$. This verifies a strictly faster uniform error bound for the sine approximation in this example.

A real or complex sequence $(a_n)$ is a Cauchy sequence ifFor $z_n=x_n+i{y}_n$, the complex modulus givesThus the same $N$ works for both coordinate Cauchy sequences. Conversely, if both real coordinate Cauchy sequences converge to $x$ and $y$, then $|z_n-(x+iy)|\le |x_n-x|+|y_n-y|\to 0$. This proves the needed completeness for complex numbers from the assumed completeness of the real numbers.

Now suppose ${\sum }_{n\ge 0}|z_n|$ converges. Its real partial sums form a Cauchy sequence. Given $\epsilon >0$, sufficiently late tails therefore satisfy ${\sum }_{k=n+1}^m|z_k|<\epsilon$. For $s_m={\sum }_{k=0}^m{z}_k$, the triangle inequality yieldsThe complex partial sums are a Cauchy sequence, hence converge by the coordinate argument. Therefore absolute convergence implies convergence for a complex series.

A series ${\sum }_{n=1}^{\infty }{a}_n$ converges if its partial sums $s_N={\sum }_{n=1}^N{a}_n$ have a finite limit of a sequence $S$ as $N\to \infty$; its sum is then $S$. It has absolute convergence if the series ${\sum }_{n=1}^{\infty }|a_n|$ converges, where $|a_n|$ is the absolute value of the term.

Assume first that the original nonnegative series converges to $S$. Its partial sums $s_m$ increase to $S$. Write $t_N={\sum }_{n=1}^N{a}_{g(n)}$ for the partial sums of the rearrangement of a series. Since $g$ is an injection, the indices $g(1),\dots ,g(N)$ are distinct, soThus $(t_N)$ is an increasing bounded sequence, and the bounded monotone sequence theorem gives $t_N\to T\le S$.

For any fixed $m$, surjectivity of $g$ givesFor $N\ge N_m$, nonnegativity implies $t_N\ge s_m$. Taking limits of sequences first in $N$ and then in $m$ gives $T\ge S$. Hence $T=S$.

Conversely, $g^{-1}$ is a bijection, and the original series is the rearrangement of a series of $\sum b_n$ defined by $g^{-1}$. The same argument proves its convergence whenever $\sum b_n$ converges. Both series converge together, and their sums are equal.

Start with all of $N_0$ and $N_1$, namely $-1,-1/2,-1/2$, whose sum is $-2$. For $n=0,1,2,\dots$ in order, append $2^{n+1}$ copies of the tripleThis uses all $2^{n+1}$ occurrences of $P_n$ and all $2^{n+2}$ occurrences of $N_{n+2}$, exactly once. Together with the initial terms, every original occurrence is used, establishing a rearrangement of a series. Its beginning isEach triple has sum zero. At its intermediate terms the partial sum is at most $2^{-(n+1)}$ above $-2$. This bound tends to zero, so the whole series, rather than just the subsequence of sums at completed triples, has sum $\boxed{-2}$.

Place the single term in $N_0$ first. For $n=0,1,2,\dots$ in order, append $2^{n+1}$ copies of the pairThis uses all of $P_n$ and all of $N_{n+1}$ exactly once, so it is an explicit rearrangement of a series. Its beginning isEvery completed pair leaves the partial sum at $-1$. Within a level-$n$ pair the partial sum differs from $-1$ by at most $2^{-(n+1)}$. As $n\to \infty$ this tends to zero, proving that the rearranged series has sum $\boxed{-1}$.

In level $n$, each group isand there are $2^n$ such groups. The partial sums at completed groups are therefore zero. Inside a group of level $n$, the only other partial sums are $2^{-(n+1)}$ and $2^{-n}$. Every level contains finitely many terms, so the level containing the $N$th term tends to infinity with $N$. Thus all partial sums lie between zero and a bound tending to zero. By the squeeze theorem, the series has sum $\boxed{0}$.

For the next two parts, label the positive occurrences of size $2^{-(n+1)}$ by $P_n$ and the negative occurrences of size $-2^{-n}$ by $N_n$. There are exactly $2^{n+1}$ occurrences in $P_n$ and $2^n$ in $N_n$. Occurrences with equal value retain separate labels, so checking their counts verifies a bijection of terms.

A series ${\sum }_{n=1}^{\infty }{a}_n$ converges when its partial sums $s_N={\sum }_{n=1}^N{a}_n$ have a finite limit of a sequence.

The comparison test says that if $0\le a_n\le b_n$ eventually and $\sum b_n$ converges, then $\sum a_n$ converges. Indeed, the partial sums of the nonnegative terms $a_n$ are increasing, and are bounded above by a fixed initial contribution plus the sum of the comparison series. The bounded monotone sequence theorem proves their convergence. Finite initial terms do not affect convergence. Applying this result to $|a_n|$ also proves absolute convergence whenever $|a_n|\le b_n$ eventually with $\sum b_n<\infty$.

Using the usual piecewise smooth interpretation of the derivative hypothesis, Dirichlet's theorem on Fourier series gives the average of the two one-sided limits of the appropriate periodic extension. The even extension is continuous at $0$ and at the identified endpoints $\pm \pi$, so the Fourier cosine series converges toFor the odd extension, the one-sided limits at zero are $-f(0)$ and $f(0)$, while at $\pi$ the periodic one-sided limits are $f(\pi )$ and $-f(\pi )$. Hence the Fourier sine series converges toAt an endpoint this equals $f$ precisely when $f$ is zero there. One can also see the zero endpoint values directly: every partial sum consists of sines that vanish at both endpoints.

The Fourier cosine series converges to the continuous even $2\pi$-periodic extension. If $r\in [-\pi ,\pi ]$ represents $x$ modulo $2\pi$, its value isIt has value zero at $x=2j\pi$, rises linearly with distance from these points until height $\pi /2$, and then stays flat until the next descending ramp. Thus the sketch on $[-3\pi ,3\pi ]$ has minima at $-2\pi ,0,2\pi$, and plateaux centred at the odd multiples of $\pi$. Its large partial sums round the corners and closely follow the whole graph.

The Fourier sine series converges, away from the odd multiples of $\pi$, to the odd $2\pi$-periodic extensionAt every odd multiple of $\pi$ its value is zero, the midpoint of the jump from $\pi /2$ to $-\pi /2$. At even multiples of $\pi$ it passes continuously through zero. Thus on every interval $((2j-1)\pi ,(2j+1)\pi )$ it has a negative plateau, a rising linear segment through $2j\pi$, and a positive plateau. The jump size is $\pi$.

The blue curves are the partial sums through $n=60$. The open endpoint circles and filled zero dots in the lower panel distinguish the one-sided limits from the value of the Fourier sine series at each jump. Near these jumps the partial sums exhibit the Gibbs phenomenon: the oscillations become concentrated in narrower regions as the number of terms grows, while their limiting overshoot remains about $0.08949$ times the jump, here about $0.281$.

The series $\sum u_n$ has absolute convergence if $\sum |u_n|$ is a convergent series, meaning its nonnegative partial sums have a finite limit. Absolute convergence implies convergence of the original real or complex series. The converse fails: the alternating harmonic series converges, while its absolute-value series is the divergent harmonic series.

A series ${\sum }_{n=1}^{\infty }{u}_n$ is a convergent series if its partial sums $S_N={\sum }_{n=1}^N{u}_n$ have a finite limit $S$ as $N\to \infty$. The number $S$ is the sum of the series. Explicitly, for every $\epsilon >0$ there is an integer $N_0$ such that $|S_N-S|<\epsilon$ whenever $N\ge N_0$. The condition $u_n\to 0$ alone is insufficient: the harmonic series has terms tending to zero but diverges.

Add the supplied five-term partial sum to each side of the tail inequality. Since $1/6=600/3600$ and $1/5=720/3600$,Numerically this is approximately $1.63028<\sum n^{-2}<1.66361$.

A convergent series ${\sum }_{n\ge 1}{u}_n$ has partial sums $S_N={\sum }_{n=1}^N{u}_n$ tending to a finite limit $S$. Explicitly, for every $\epsilon >0$ there is an integer $N_0$ such that $|S_N-S|<\epsilon$ whenever $N\ge N_0$. The convergence of the individual terms to zero is necessary but is not sufficient for convergence of the series.

For $x\in (n,n+1)$, strict decrease gives $f(x)<f(n)$, while for $x\in (n-1,n)$ it gives $f(x)>f(n)$. Integrating over these unit intervals provesSumming the rectangle bounds givesThe partial sums and the improper integral are nondecreasing. These bounds show that one is bounded exactly when the other is bounded, proving the integral test for convergence. In the convergent case, sum the same interval inequalities for $n=N+1,N+2,\dots$ to obtainFor $f(x)=1/(x\ln x)$, the antiderivative is $\ln (\ln x)$, which grows without bound, so the first series diverges. For $f(x)=1/[x(\ln x{)}^2]$, the antiderivative is $-1/\ln x$, whose tail integral is finite, so the second series converges, withBoth functions are positive and strictly decreasing on $[2,\infty )$, as required.

With the standard real normalization, the Fourier series of a period-$2L$ function iswhereFor the partial sum $S_N$, Fourier orthogonality eliminates all cross terms. The constant term contributes $2L(a_0/2{)}^2$, while every squared sine or cosine integrates to $L$, givingFor a square-integrable function, completeness of the trigonometric orthogonal basis in the Hilbert space $L^2[-L,L]$ makes $S_N\to f$ in norm. Passing to the limit proves Parseval's identity:Square integrability is the needed hypothesis here; periodicity alone is not an integrability or convergence condition.