Rearranging an absolutely convergent series keeps its sum; a conditionally convergent series can be rearranged to add up to any number you like (Riemann).
Esta lección aún no está traducida. Por ahora se muestra en inglés; la versión en español está en camino.
For finite sums, the order of the terms never matters. For series it can. Whether it does depends on the difference between absolute and conditional convergence, and the alternating harmonic series is the standard example.
🔑 Idea clave
If $\sum a_k$ converges absolutely, every rearrangement converges to the same sum. If $\sum a_k$ converges conditionally, then for every real number $T$ some rearrangement converges to $T$ (Riemann's rearrangement theorem), and others diverge.
💡 Por qué funciona
Conditional case. Split the terms into positives and negatives. Both parts must diverge: if one converged, the other would equal the original series minus it and converge too, making the series absolutely convergent. So we have an unlimited supply of positive terms and of negative terms, each going to $0$. To reach $T$: add positive terms until the sum first exceeds $T$, then negative terms until it first drops below $T$, and repeat. Each overshoot is at most the last term used, and those terms go to $0$, so the partial sums converge to $T$. Absolute case. Given $\varepsilon > 0$, choose $N$ with $\sum_{k > N} |a_k| < \varepsilon$. Far enough out, a rearrangement has used all of $a_1, \ldots, a_N$, and what remains is made of terms beyond $N$, whose total size is less than $\varepsilon$. So the rearranged sums end up within $\varepsilon$ of the original sum.
A famous case: taking two positive terms, then one negative, $1 + \frac{1}{3} - \frac{1}{2} + \frac{1}{5} + \frac{1}{7} - \frac{1}{4} + \cdots$, gives $\frac{3}{2}\ln 2$ instead of $\ln 2$. In general, $p$ positive terms then $q$ negative terms give $\ln 2 + \frac{1}{2}\ln\frac{p}{q}$.
The alternating harmonic series in its original order settles at $\ln 2 \approx 0.693$. A different order of the same terms settles somewhere else.
Worked examples
✏️ Ejemplo resuelto: A rearranged partial sum
Find the sum of the first $6$ terms of $1 + \frac{1}{3} - \frac{1}{2} + \frac{1}{5} + \frac{1}{7} - \frac{1}{4} + \cdots$.
✏️ Ejemplo resuelto: Riemann's method for a target
To rearrange $1 - \frac{1}{2} + \frac{1}{3} - \cdots$ toward $T = \frac{3}{2}$, how many positive terms come first?
Running sums: $1$, then $1.333$, then $1.533$.
The sum first exceeds $1.5$ after three positive terms.
Respuesta: $3$; then negative terms are added until the sum drops below $1.5$ again.
⚠️ Error común
Assuming addition is always commutative. Rearranging the terms of a conditionally convergent series can change its sum, or make it diverge. Before reordering or regrouping an infinite sum in a proof, check absolute convergence: for $\sum \frac{(-1)^{k+1}}{k^{2}}$ any order gives $\frac{\pi^{2}}{12}$, for $\sum \frac{(-1)^{k+1}}{k}$ it does not.
Práctica
Apruebas con 3 respuestas correctas, las 2 últimas seguidas. Cada respuesta correcta da 3 XP.