Infinite Series

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AP Calculus BC Topic 10 8:20 English narration · English + 中文 subtitles burned in

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Can you add infinitely many numbers and get a finite answer? 你能把无穷多个数相加，却得到一个有限的答案吗？

Surprisingly, yes. 令人惊讶的是，可以。

Add forever, and sometimes the total runs away to infinity — but sometimes it settles down. 一直加下去，有时总和会跑向无穷——但有时它会稳定下来。

Take one half, then a quarter, then an eighth, each time halving the step. 取二分之一，再取四分之一，再取八分之一，每次把步长减半。

You never overshoot, and the running total settles on exactly one. 你永远不会超过，而累计的总和恰好稳定在一。

That is a convergent series — and this unit is about telling which sums settle, and which do not. 这就是一个收敛级数——而本单元讲的就是如何判断哪些和会稳定，哪些不会。

Today we build the whole series toolkit — the tests for convergence, and then Taylor series, which turn functions into infinite polynomials. 今天我们来建立整套级数工具——各种收敛判别法，然后是泰勒级数，它把函数变成无穷次多项式。

Let's begin. 让我们开始吧。

First, what does an infinite sum even mean? 首先，无穷和到底是什么意思？

We define it as the limit of its partial sums. 我们把它定义为部分和的极限。

Add the first term, then the first two, then the first three — each partial sum climbs a little closer. 先加第一项，再加前两项，再加前三项——每个部分和都更靠近一点。

If those partial sums approach a finite number, the series converges to it. 如果这些部分和趋近一个有限的数，级数就收敛到它。

If they do not, it diverges. 如果不趋近，级数就发散。

So every convergence question is really a question about a limit. 所以每一个收敛性问题，其实都是一个极限问题。

One family we can sum exactly: the geometric series, where each term is the previous one times a constant ratio. 有一族级数我们可以精确求和：几何级数，它的每一项都是前一项乘以一个固定的公比。

It converges exactly when the size of the ratio is less than one — and then the sum is the first term, divided by one minus the ratio. 当公比的绝对值小于一时，它恰好收敛——此时和等于首项除以一减公比。

For three, plus three halves, plus three quarters, and so on, the ratio is one half, so the sum is six. 对三，加二分之三，加四分之三，如此下去，公比是二分之一，所以和是六。

This is the only series you can sum on sight, and it underlies power series later. 这是唯一一个你能一眼求和的级数，而它也是后面幂级数的基础。

Two facts you must know. 有两个必须掌握的事实。

First, the nth-term test: if the terms do not shrink to zero, the series must diverge. 第一，n 项判别法：如果各项不趋于零，级数一定发散。

But careful — this is only a test for divergence. 但要小心——这只是发散的判别法。

If the terms do go to zero, it tells you nothing. 如果各项确实趋于零，它什么也说明不了。

The proof is the harmonic series: one, plus a half, plus a third, and so on. 证据就是调和级数：一，加二分之一，加三分之一，如此下去。

Its terms go to zero, yet the sum still diverges. 它的各项趋于零，但和仍然发散。

More generally, the p-series converges when p is bigger than one, and diverges otherwise. 更一般地，p 级数在 p 大于一时收敛，否则发散。

That family is our measuring stick. 这一族就是我们的标尺。

Now the toolkit — pick the right tool for the series in front of you. 现在讲工具箱——为眼前的级数挑选合适的工具。

The integral test turns the sum into an improper integral: both converge or both diverge together. 积分判别法把和变成一个反常积分：两者同敛同散。

Comparison tests measure an unfamiliar series against a known one, a p-series or a geometric series. Direct comparison bounds it above or below; limit comparison takes the limit of the ratio of the terms, and if that limit is finite positive, both series do the same thing. 比较判别法把一个陌生的级数与已知的级数作比较，比如 p 级数或几何级数—— 可以直接比较，也可以取两者各项之比的极限。

The alternating series test handles signs that flip: if the sizes are decreasing and go to zero, it converges. 交错级数判别法处理符号交替的情形：只要各项的大小递减趋于零，它就收敛。

And the ratio test is the workhorse for factorials and nth powers. 而比值判别法是处理阶乘和 n 次幂的主力。

Let's run the ratio test. 我们来用一次比值判别法。

Test the sum of n over two to the n. 检验 n 除以二的 n 次方的和。

Take the size of the ratio of consecutive terms: n plus one, over two to the n plus one, times two to the n, over n. 取相邻两项之比的绝对值：n 加一，除以二的 n 加一次方，再乘以二的 n 次方除以 n。

The powers of two cancel down to a single half, leaving n plus one over two n. 二的幂约掉，只剩一个二分之一，留下二 n 分之 n 加一。

As n grows, that approaches one half, which is less than one — so the series converges. 当 n 增大时，它趋近二分之一，小于一——所以级数收敛。

Remember: less than one converges, greater than one diverges, and exactly one is inconclusive — the test tells you nothing. 记住：小于一收敛，大于一发散，恰好等于一则无法判断。

There are two grades of convergence. 收敛分两个等级。

A series converges absolutely if the sum of the sizes of the terms already converges — the signs are not doing any work. 如果各项绝对值之和本身就收敛，那么级数绝对收敛——符号并没有起作用。

It converges conditionally if the series converges, but the sum of the sizes diverges — so the convergence depends entirely on cancellation between plus and minus. 如果级数收敛，但各项绝对值之和发散，那么它条件收敛—— 也就是说收敛完全依赖于正负之间的相互抵消。

The classic example is the alternating harmonic series: it converges, but strip the signs away and you are back to the divergent harmonic series. 经典例子是交错调和级数：它收敛，但把符号去掉，你就回到了发散的调和级数。

Now the second half of the unit — and one of the most beautiful ideas in calculus. 现在进入本单元的后半部分——也是微积分中最优美的思想之一。

Take a function, pick a centre point, and build a polynomial that matches its value there, then its slope, then its curvature, and so on. 取一个函数，选一个中心点，构造一个多项式：先让它在该点的函数值相符，再让斜率相符，再让弯曲程度相符，如此下去。

Each extra term matches one more derivative, and the polynomial hugs the curve over a wider and wider range. 每多一项，就多匹配一阶导数，而这个多项式就在越来越宽的范围内紧贴曲线。

Watch how the fit improves with every term. 看每加一项，拟合是如何变好的。

Here is the formula. 公式是这样的。

The Taylor polynomial of degree n, centred at a, is the sum of the kth derivative at a, divided by k factorial, times x minus a to the k. 以 a 为中心、n 次的泰勒多项式，等于对 k 求和：a 处的 k 阶导数，除以 k 的阶乘，再乘以 x 减 a 的 k 次方。

Every term matches one more derivative at the centre, so the fit is best near the centre and worsens as you move away. 每一项都多匹配函数在中心处的一阶导数。

When the centre is zero, we call it a Maclaurin polynomial. 当中心取零时，我们称它为麦克劳林多项式。

A common exam task: build the polynomial from a table of derivative values, then use it to estimate. 考试常见任务是：从一张导数值表构造这个多项式，再用它来估值。

Extend to infinitely many terms and you get a Taylor series. 把项数扩展到无穷，就得到泰勒级数。

There are four you must memorize: e to the x, sine, cosine, and one over one minus x — which is just our geometric series again. 有四个你必须背下来：e 的 x 次方、正弦、余弦，以及一减 x 分之一——后者其实就是我们的几何级数。

From these four you build new ones by manipulating them, without differentiating anything: substitute, or the series can be differentiated and integrated term by term. 有了这四个，你就能构造新的级数，而不必求任何导数。

Want the series for e to the x squared? 想要 e 的 x 平方次方的级数？

Just substitute x squared into the series for e to the x. 只要把 x 平方代入 e 的 x 次方的级数即可。

That trick is far faster than differentiating six times. 这个技巧比求六次导数快得多。

Substituting is one move; the other two are differentiating and integrating, and inside the radius you may do either one term by term. 代入是一种手法；另外两种是求导和积分，而在收敛半径之内，你可以逐项做这两件事中的任何一件。

Watch how much that buys you. 看看这能带来多少东西。

Start from the geometric series: one over one minus x is the sum of x to the n. 从几何级数出发：一比一减 x 等于 x 的 n 次方之和。

Now integrate both sides. 现在两边同时积分。

On the right, integrating x to the n term by term gives x to the n plus one, over n plus one. 右边逐项对 x 的 n 次方积分，得到 x 的 n 加一次方，比 n 加一。

On the left, the integral of one over one minus x is minus the natural logarithm of one minus x. 左边，一比一减 x 的积分是负的一减 x 的自然对数。

So we have just derived a series for a logarithm, using no derivatives at all. 于是我们完全没有求任何导数，就推出了一个对数函数的级数。

One caution: the radius of convergence survives the operation, but the behaviour at the ends can change, so retest the endpoints every time. 有一点要当心：收敛半径在这种运算下不变，但端点处的表现可能改变，所以每次都要重新检验端点。

And note what we have built, because the BC series free-response is almost always these three moves in order: derive a new series from a known one, find its interval of convergence, then bound the error of a partial sum. 还要注意我们刚才做了什么，因为 BC 的级数自由作答题几乎总是按顺序做这三步：从已知级数推出新级数，求它的收敛区间，然后估计部分和的误差界。

An approximation is only useful if you can bound its error. 一个近似只有在你能界定它的误差时才有用。

For a convergent alternating series there is a beautifully simple rule: the error is no larger than the first omitted term. 对收敛的交错级数，有一条极其简单的规则：误差不会超过你略去的第一项。

For a Taylor polynomial in general, use the Lagrange error bound: the maximum of the next derivative on the interval, divided by n plus one factorial, times the distance from the centre, to that power. 而对一般的泰勒多项式，就用拉格朗日误差界：下一阶导数在该区间上的最大值，除以 n 加一的阶乘，再乘以到中心的距离的相应次方。

Bound that derivative, then compute. 先界定那个导数，再计算。

Finally, a power series does not converge everywhere. 最后，幂级数并非处处收敛。

It converges within a radius of convergence, R, of its centre. 它在以中心为心、半径为 R 的范围内收敛。

Find it with the ratio test: set the limit less than one and solve. 用比值判别法求出它：令极限小于一并求解。

But the ratio test says nothing at the edges, so you must test the two endpoints separately, one at a time. 但比值判别法在边界处什么也说明不了，所以你必须把两个端点分别单独检验。

For the sum of x to the n over n, the radius is one; at minus one you get the convergent alternating harmonic series, and at plus one the divergent harmonic series — so the interval includes minus one but excludes one. 对 x 的 n 次方除以 n 的和，半径是一；在负一处得到收敛的交错调和级数，在正一处得到发散的调和级数—— 所以区间包含负一，但不包含一。

Before you go, three marks to keep. 结束之前，三个要守住的分。

First, terms going to zero never proves convergence — the harmonic series is the counterexample. 第一，各项趋于零绝不能证明收敛——调和级数就是反例。

Second, after the ratio test gives you a radius, always test the endpoints separately; the interval is not complete without them. 第二，比值判别法给出半径之后，一定要分别检验两个端点；没有它们，区间就不完整。

Third, to build a new Maclaurin series, start from a series you know and substitute, differentiate, or integrate — do not grind out derivatives. 第三，要构造新的麦克劳林级数，就从你已知的级数出发，做代入、求导或积分—— 不要硬算导数。

This topic is yours. 这个专题就是你的了。