具體數學--(無限數列和)

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Infinite sum

examples

1. The infinite sum
   $S=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+...$
   is equal to 2, because if we double it we get
   $2S=2+1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+...=2+S$

2. However, if we look at $T=1+2+4+8+16+32+...$, we will get weird
   result from previous method.

   $2T=2+4+8+16+32+...=T-1$

   It means $T=-1$ which not correct.

In order to avoid such incorrect result, we need a better definition for
$\sum_{k\in K}a_k$, where $K$ might be infinite.

Definition

type I $\sum_{k\in F}a_k\leqslant A$

If there is a bouding constant A, we define that $\sum_{k\in K}a_k$ to
be the least such A.Otherwise, $\sum_{k\in K}a_k=\infty$.

Now we can define

$\sum_{k\geqslant 0}x^k=\lim_{n\rightarrow \infty}\frac{1-x^{n+1}}{1-x}=\left\{ \begin{aligned} 1/(1-x),& (0\leqslant x<1)\\ \infty,&x\geqslant 1 \end{aligned} \right.$

However, there still some weird sums such as

$\sum_{k\geqslant 0}(-1)^k=1-1+1-1+1-1+...$

if we group the terms in pairs, we get

$\left. \begin{aligned} (1-1)+(1-1)+(1-1)+...=0+0+0+...=0\\ 1-(1-1)-(1-1)-(1-1)-...=1-0-0-0-...=1 \end{aligned} \right.$

Although, we have the following fact.
$\sum_{k\geqslant 0}(-1)^k \leqslant 1$

Another
example,$...+(-\frac{1}{4})+(-\frac{1}{3})+(-\frac{1}{2})+1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...$

$\left. \begin{aligned} \cdots+\left(-\frac{1}{4}+\left(-\frac{1}{3}+\left(-\frac{1}{2}+(1)+\frac{1}{2}\right)+\frac{1}{3}\right)+\frac{1}{4}\right)+\cdots&=1\\ \cdots+\left(-\frac{1}{4}+\left(-\frac{1}{3}+\left(-\frac{1}{2}+1+\frac{1}{2}\right)+\frac{1}{3}+\frac{1}{4}\right)+\frac{1}{5}+\frac{1}{6}\right)+\cdots&=1+H_{2n}-H_{n+1},(\lim_{x\rightarrow \infty}(H_{2n}-H_{n+1})=ln2) \end{aligned} \right.$

type II

$x=x^+-x^-,x^+=x[x>0],x^-=-x[x<0]$

Our general definition is

$\sum_{k\in K}a_k=\sum_{k\in K} a_k^+-\sum_{k\in K}a_k^-$

Now, let’s define

$A^+=\sum_{k\in K}a_k^+,A^-=\sum_{k\in K}a_k^-$

• if $A^+$ and $A^-$ are both finite, the sum $\sum_{k\in K} a_k$ is
  said to converge absolutely to the value $A=A^+-A^-$

• If $A^+=\infty$ but $A^-$ is finite, the sum $\sum_{k\in K}a_k$ is
  said to diverge to $+\infty$

• If $A^-=\infty$ but $A^+$ is finite, the sum $\sum_{k\in K}a_k$ is
  said to diverge to $-\infty$

• If $A^+=A^-=\infty$, all things are meaningless.

We also can extend this defintion to complex domain.
$\sum_{k\in K}R_{a_{k}}+i\sum_{k\in K}I_{a_{k}}$

If the real and img part are both defined,we say that the complex sum
defined. Otherwise, undefined.

multiple indexs

$\sum_{j\in J, k\in K_j}a_{j,k}$ converges absolutely to A, then there
exist complex numbers $A_j$ for each $j\in J$ such that

• $\sum_{k\in K_j} a_{j,k}$ converges absolutely to $A_j$, and

• $\sum_{j\in J}A_{j}$ converges absolutely to A.
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