Working with Geometric Series · 处理几何级数
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| geometric series/ˌdʒiːəʊˈmetrɪk ˈsɪəriːz/ | 几何级数 | jǐ hé jí shù |
| common ratio/ˈkɒmən ˈreɪʃɪəʊ/ | 公比 | gōng bǐ |
The one series you can sum exactly
- Most series are hard to add up. The geometric series 几何级数 is the friendly exception — it has a formula.
- Each term is the previous one times a fixed common ratio 公比 $r$: $a+ar+ar^2+ar^3+\cdots$.
- It's the infinite version of the geometric sequences you already know.
- And there's a clean rule for exactly when it converges and to what.
唯一能精确求和的级数
- 大多数级数很难加起来。几何级数是友好的例外——它有公式。
- 每一项是前一项乘以固定的公比 $r$:$a+ar+ar^2+ar^3+\cdots$。
- 它是你已知的等比数列的无穷版本。
- 而且对它何时收敛、收敛到什么,有一条干净的规则。
A geometric sequence · 几何数列
A geometric series adds terms multiplied by $r$ each time — it converges (to $a/(1-r)$) exactly when $|r|<1$. · 几何级数将各项乘以 $r$ 每次 — 它收敛于（到 $a/(1-r)$）当且仅当 $|r|<1$.
Converges only when $|r|<1$
- A geometric series converges exactly when $|r|<1$ (the ratio is a fraction).
- If $|r|\ge1$, the terms don't shrink to $0$, so it diverges.
- $\tfrac12+\tfrac14+\tfrac18+\cdots$ ($r=\tfrac12$) converges; $1+2+4+\cdots$ ($r=2$) diverges.
- The single number $|r|$ decides everything.
仅当 $|r|<1$ 才收敛
- 几何级数收敛当且仅当 $|r|<1$(公比是个分数)。
- 若 $|r|\ge1$,各项不缩向 $0$,所以发散。
- $\tfrac12+\tfrac14+\tfrac18+\cdots$($r=\tfrac12$)收敛;$1+2+4+\cdots$($r=2$)发散。
- 单个数 $|r|$ 决定一切。
A geometric series $\sum a r^n$ converges exactly when... · 几何级数$\sum a r^n$当且仅当...时收敛。
Converges iff the ratio is a fraction, $|r|<1$. · 当且仅当比值为分数$|r|<1$时收敛。
Which geometric series diverges? · 哪个几何级数发散？
$r=2$ has $|r|\ge1$ → diverges. · $r=2$具有$|r|\ge1$ → 发散。
The sum formula
- When it converges ($|r|<1$), the sum is
-
$$\sum_{n=0}^{\infty} a\,r^{n}=\frac{a}{1-r}$$
- $a$ is the first term and $r$ the common ratio.
- Plug in and you get the exact infinite sum — no limits to compute by hand.
求和公式
- 当它收敛($|r|<1$),和是
-
$$\sum_{n=0}^{\infty} a\,r^{n}=\frac{a}{1-r}$$
- $a$ 是首项,$r$ 是公比。
- 代入即得精确的无穷和——不用手算极限。
The sum of a convergent geometric series is... · 收敛几何级数的和为...
$\sum a r^n=\frac{a}{1-r}$ for $|r|<1$. · $\sum a r^n=\frac{a}{1-r}$ (针对$|r|<1$)。
The formula $\tfrac{a}{1-r}$ can be applied even when $|r|\ge1$. · 公式$\tfrac{a}{1-r}$即使在$|r|\ge1$的情况下也可应用。
Only when $|r|<1$ (convergent). · 仅当$|r|<1$ (收敛) 时适用。
Getting $a$ and $r$ right
- $a$ is the actual first term of the series (whatever it starts at), not necessarily $1$.
- $r$ is the ratio between consecutive terms — divide any term by the one before it.
- If a series starts at $n=1$ instead of $n=0$, adjust $a$ to be that first term.
- With the right $a$ and $r$, the formula $\tfrac{a}{1-r}$ does the rest.
把 $a$ 和 $r$ 求对
- $a$ 是级数的实际首项(无论从什么开始),不一定是 $1$。
- $r$ 是相邻项之比——用任一项除以它前一项。
- 若级数从 $n=1$ 而非 $n=0$ 开始,把 $a$ 调整为那个首项。
- 有了正确的 $a$ 和 $r$,公式 $\tfrac{a}{1-r}$ 搞定其余。
Find $\displaystyle\sum_{n=0}^\infty 5\left(\tfrac13\right)^n$ (a decimal). · 求$\displaystyle\sum_{n=0}^\infty 5\left(\tfrac13\right)^n$ (小数形式)。
$\frac{5}{1-1/3}=\frac{5}{2/3}=7.5$.
In the formula $\tfrac{a}{1-r}$, $a$ is the ____ term of the series. · 在公式$\tfrac{a}{1-r}$中，$a$是级数的____项。
$a$ is the actual first term. · $a$是实际的第一个项。
The formula $\frac{a}{1-r}$ applies only when $|r|<1$ — using it on a divergent geometric series gives a meaningless number. And $a$ is the first term as written, not always $1$: for $\sum_{n=1}^\infty 3\cdot 2^{-n}=\tfrac32+\tfrac34+\cdots$, the first term is $\tfrac32$, not $3$.
公式 $\frac{a}{1-r}$ 仅在 $|r|<1$ 时适用——把它用在发散的几何级数上会得到无意义的数。而 $a$ 是写出来的首项,不总是 $1$:对 $\sum_{n=1}^\infty 3\cdot 2^{-n}=\tfrac32+\tfrac34+\cdots$,首项是 $\tfrac32$,不是 $3$。
Find the sum of $\displaystyle\sum_{n=0}^{\infty} 5\left(\tfrac13\right)^{n}$.
- First term $a=5$, common ratio $r=\tfrac13$. Since $|r|<1$, it converges.
- $\displaystyle\sum=\frac{a}{1-r}=\frac{5}{1-\tfrac13}=\frac{5}{\tfrac23}=\frac{15}{2}$.
求 $\displaystyle\sum_{n=0}^{\infty} 5\left(\tfrac13\right)^{n}$ 的和。
- 首项 $a=5$,公比 $r=\tfrac13$。因为 $|r|<1$,它收敛。
- $\displaystyle\sum=\frac{a}{1-r}=\frac{5}{1-\tfrac13}=\frac{5}{\tfrac23}=\frac{15}{2}$。
A geometric series $\sum a\,r^n$ converges iff $|r|<1$, and then sums to $\frac{a}{1-r}$ (with $a$ the first term, $r$ the common ratio). If $|r|\ge1$ it diverges. It's the one series with an exact, easy sum.
几何级数 $\sum a\,r^n$ 收敛当且仅当 $|r|<1$,此时和为 $\frac{a}{1-r}$($a$ 是首项,$r$ 是公比)。若 $|r|\ge1$ 则发散。它是唯一有精确、易求和的级数。