Summation of Series by the Method of Differences | 差分法求和

📚 Summation of Series by the Method of Differences | 差分法求和

The method of differences is a powerful technique for summing finite series whose terms can be expressed as the difference of two consecutive terms of another sequence. Often called “telescoping”, this method simplifies a long summation into just a few evaluations at the boundaries.

差分法是求有限级数和的一种强大技巧，适用于项可以表示为另一个数列相邻两项之差的级数。它通常被称为“裂项相消”，能将长求和简化为仅在边界处的若干计算。

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1. The Core Idea | 核心思想

Suppose we can write the general term \(u_k\) of a series as \(u_k = v_{k+1} – v_k\) for some sequence \(v_k\). Then the sum from \(k=1\) to \(n\) becomes:

假设我们能将级数的通项 \(u_k\) 写成 \(u_k = v_{k+1} – v_k\)，其中 \(v_k\) 是某个数列。则从 \(k=1\) 到 \(n\) 的和变为：

∑_{k=1}^{n} (v_{k+1} – v_k) = v_{n+1} – v_1

All intermediate terms cancel in pairs. This is why the method is also called “telescoping” – the sum collapses like a telescope.

所有中间项成对相消。这也是为什么该方法也叫“望远镜求和”——求和如同望远镜一样收缩。

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2. When to Use It | 何时使用

The method is particularly effective when the general term is a rational function whose denominator factors into consecutive integers, or when it can be expressed as a partial fraction difference. Typical forms include:

当通项是分母可分解为连续整数的有理函数，或者可以表示为部分分式之差时，该方法尤为有效。常见形式包括：

• \(\frac{1}{k(k+1)}\) – decomposes to \(\frac{1}{k} – \frac{1}{k+1}\)
• \(\frac{1}{k(k+1)(k+2)}\) – can be split using partial fractions
• \(\frac{1}{(2k-1)(2k+1)}\) – useful for series of reciprocals of odd products

• \(\frac{1}{k(k+1)}\) – 分解为 \(\frac{1}{k} – \frac{1}{k+1}\)
• \(\frac{1}{k(k+1)(k+2)}\) – 可用部分分式拆分
• \(\frac{1}{(2k-1)(2k+1)}\) – 适用于奇数乘积倒数级数

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3. Partial Fractions and Differences | 部分分式与差分

Often the given term is not already in the form \(v_{k+1} – v_k\). We first carry out a partial fraction decomposition. For example:

通常给出的项并不直接是 \(v_{k+1} – v_k\) 的形式，我们需要先进行部分分式分解。例如：

\(\frac{1}{k(k+2)} = \frac{1}{2}\left( \frac{1}{k} – \frac{1}{k+2} \right)\)

This introduces a “gap” of 2 between the terms, so when summing, terms do not all cancel; only every second term cancels, leaving two boundary sums.

这里出现了差距为2的项，因此在求和时并非所有项都相消，而是每隔一项相消，剩余两个边界和。

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4. Worked Example 1: Basic Telescoping | 例1：基础裂项

Find the sum \(S_n = \sum_{k=1}^{n} \frac{1}{k(k+1)}\).

求 \(S_n = \sum_{k=1}^{n} \frac{1}{k(k+1)}\) 的和。

We decompose the term:

我们对通项进行分解：

\(\frac{1}{k(k+1)} = \frac{1}{k} – \frac{1}{k+1}\)

Then the sum is:

于是和为：

\(S_n = \left(1 – \frac{1}{2}\right) + \left(\frac{1}{2} – \frac{1}{3}\right) + \cdots + \left(\frac{1}{n} – \frac{1}{n+1}\right) = 1 – \frac{1}{n+1}\)

Hence \(S_n = \frac{n}{n+1}\).

因此 \(S_n = \frac{n}{n+1}\)。

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5. Worked Example 2: Non-unit Numerator | 例2：非单位分子

Sum the series \(\sum_{k=1}^{n} \frac{2}{k(k+2)}\).

求级数 \(\sum_{k=1}^{n} \frac{2}{k(k+2)}\) 的和。

Decompose:

分解：

\(\frac{2}{k(k+2)} = \frac{1}{k} – \frac{1}{k+2}\)

Now write out the terms:

现在写出各项：

\(S_n = \left(1 – \frac{1}{3}\right) + \left(\frac{1}{2} – \frac{1}{4}\right) + \left(\frac{1}{3} – \frac{1}{5}\right) + \cdots\)

After cancellation, the remaining terms are \(1 + \frac{1}{2} – \frac{1}{n+1} – \frac{1}{n+2}\). Therefore:

相消后，剩余项为 \(1 + \frac{1}{2} – \frac{1}{n+1} – \frac{1}{n+2}\)。因此：

\(S_n = \frac{3}{2} – \frac{1}{n+1} – \frac{1}{n+2}\)

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6. Systematic Cancellation Table | 系统化的相消表

For more complex decompositions, it helps to list the first few and the last few terms explicitly. Consider a gap of \(d\): the remaining terms are the first \(d\) terms of \(v_k\) and the last \(d\) terms of \(v_k\).

对于更复杂的分解，明确列出前几项和后几项会有帮助。设差距为 \(d\)，则剩余项为 \(v_k\) 的前 \(d\) 项和后 \(d\) 项。

Gap d	Remaining positive terms	Remaining negative terms
1	\(v_1\)	\(v_{n+1}\)
2	\(v_1 + v_2\)	\(v_{n+1} + v_{n+2}\)
3	\(v_1 + v_2 + v_3\)	\(v_{n+1} + v_{n+2} + v_{n+3}\)

Always verify the pattern by writing three or four terms explicitly before jumping to the final formula.

在直接写出最终公式前，务必先写出三四项来验证规律。

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7. When the Term Is Not Already a Difference | 当通项本身不是差的形式

Sometimes the term \(u_k\) can be written as \(v_k – v_{k+1}\) (reverse order) or as \(v_k – v_{k+m}\) with a constant \(m\). In such cases, the same telescoping logic applies, but careful attention is needed to the sign and the boundaries.

有时通项 \(u_k\) 可以写成 \(v_k – v_{k+1}\)（反向）或 \(v_k – v_{k+m}\)（其中 \(m\) 为常数）。此时同样的裂项逻辑适用，但需要仔细注意符号和边界。

For example:

例如：

\(\frac{1}{(2k-1)(2k+1)} = \frac{1}{2}\left( \frac{1}{2k-1} – \frac{1}{2k+1} \right)\)

Here \(v_k = \frac{1}{2k-1}\), and the gap is 2. The sum is:

这里 \(v_k = \frac{1}{2k-1}\)，差距为2。其和为：

\(S_n = \frac{1}{2}\left( 1 – \frac{1}{2n+1} \right)\)

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8. Combining with Standard Formulas | 与标准公式结合

The method of differences can also be used to derive summation formulas for polynomial sequences. For instance, since \((k+1)^3 – k^3 = 3k^2 + 3k + 1\), summing from 1 to n gives a direct proof of the sum of squares formula.

差分法也可用于推导多项式数列的求和公式。例如，因为 \((k+1)^3 – k^3 = 3k^2 + 3k + 1\)，从1到n求和即可直接证明平方和公式。

Summing both sides:

两边求和：

\((n+1)^3 – 1 = 3\sum_{k=1}^{n} k^2 + 3\sum_{k=1}^{n} k + n\)

Rearranging gives:

整理得到：

\(\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}\)

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9. Common Pitfalls | 常见陷阱

Students often make errors in the constant factor obtained from partial fractions. For example, writing \(\frac{1}{k(k+2)} = \frac{1}{k} – \frac{1}{k+2}\) is incorrect; the correct decomposition includes a factor of \(\frac{1}{2}\).

学生常在部分分式中的常数因子出错。例如，把 \(\frac{1}{k(k+2)} = \frac{1}{k} – \frac{1}{k+2}\) 写成这样是错误的；正确的分解包含因子 \(\frac{1}{2}\)。

Another common mistake is failing to identify which terms actually cancel. Always write out at least five terms: the first two and the last two, to see the pattern clearly.

另一个常见错误是未能识别哪些项真正相消。务必至少写出五项：前两项和后两项，以看清规律。

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10. Practice Questions | 练习

Try the following on your own:

请自行尝试以下题目：

1. \(\sum_{k=1}^{n} \frac{1}{(3k-1)(3k+2)}\) – hint: decompose into a difference with gap 3.
2. \(\sum_{k=1}^{n} \frac{k}{(k+1)!}\) – hint: express \(k = (k+1)-1\) and use factorial differences.
3. \(\sum_{k=1}^{n} \ln\left(1+\frac{1}{k}\right)\) – use logarithm laws to telescope.

1. \(\sum_{k=1}^{n} \frac{1}{(3k-1)(3k+2)}\) – 提示：分解为差距为3的差。
2. \(\sum_{k=1}^{n} \frac{k}{(k+1)!}\) – 提示：将 \(k = (k+1)-1\) 并利用阶乘差分。
3. \(\sum_{k=1}^{n} \ln\left(1+\frac{1}{k}\right)\) – 利用对数运算法则裂项。

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11. Summary | 总结

The method of differences is a cornerstone of A-level series summation. Master it by practicing partial fraction decomposition and by meticulously tracking which terms survive.

差分法是A-Level级数求和的基石。通过练习部分分式分解并仔细追踪剩余项来掌握它。

Remember the three essential steps:

记住三个关键步骤：

• Decompose the general term into a difference of two terms.
• Write out the sum explicitly for the first few and last few terms.
• Cancel matching pairs and simplify the boundary terms.

• 将通项分解为两项之差。
• 明确写出前几项和后几项的求和。
• 消去成对项并化简边界项。

With practice, telescoping sums become intuitive and quick.

多加练习，裂项求和将变得直观而快速。

Published by TutorHao | Mathematics Revision Series | aleveler.com

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