Most terms cancel

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A telescoping series is one which, when we expand it (write out a few terms), we can easily recognize that, when "re-punctuated,"" most of the "interior" terms subtract to zero, leaving just a few non-zero terms that are fairly easy to evaluate. Telescoping series are one of just a few infinite series for which we can easily calculate the sum.

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We'll begin by decomposing the general term into a sum of fractions. You can learn more about this in the section on rational decomposition, but we'll go through it in detail here, too.

We should be able to expand the general term into two fractions like this:

$$\frac{1}{n(n + 1)} = \frac{A}{n} + \frac{B}{n + 1}$$

All we have to do is find A and B. We do that like this. First, multiply both sides of the equation above by $n(n + 1).$

$$1 = A\,(n + 1) + B \, n$$

Now notwithstanding the limits of our series, this expression should be good for any n, so we can manipulate the value of n to find A and B:

$$ \begin{align} \text{Let n = 0:} \; \text{then} \; A &= 1 \\[4pt] \text{Let n = -1:} \; \text{then} \; B &= -1 \end{align}$$

So we can re-express our series like this:

$$\sum_{n = 1}^{\infty} \frac{1}{n (n + 1)} = \sum_{n = 1}^{\infty} \left( \frac{1}{n} - \frac{1}{n + 1} \right)$$

Now let's expand this for a few terms and look for a pattern as $n \to \infty.$

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$$ \begin{align} &\lim_{N\to\infty} \sum_{n = 1}^N \left( \frac{1}{n} - \frac{1}{n + 1} \right) \\[5pt] &= \lim_{N\to\infty} \left[ \left( 1 - \frac{1}{2} \right) + \left( \frac{1}{2} - \frac{1}{3} \right) + \left( \frac{1}{3} - \frac{1}{4} \right) + \dots + \left( \frac{1}{N} - \frac{1}{N + 1} \right) \right] \\[5pt] &= \lim_{N\to\infty} \left[ 1 + \left( -\frac{1}{2} - \frac{1}{2} \right) + \left( -\frac{1}{3} + \frac{1}{3} \right) + \left( -\frac{1}{4} + \frac{1}{4} \right) + \dots + \left( -\frac{1}{N} + \frac{1}{N} \right) - \frac{1}{N + 1}\right] \\[5pt] &= \lim_{N\to\infty} \left[ 1 - \frac{1}{N + 1} \right] = 1 - 0 = \bf 1 \end{align}$$

Notice that in the third line above, in which we just re-grouped the terms, all of the interior parentheses vanish (subtract to zero), and we're left with just the first and last term. The sum of this series is 1.

Example 1

Find the sum of the series $\sum_{n = 1}^{\infty} \frac{1}{(2n - 1)(2n + 1)}.$

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Solution: First decompose the general term to two fractions:

$$\frac{1}{(2n - 1)(2n + 1)} = \frac{A}{2n - 1} + \frac{B}{2n + 1}$$

Multiplying both sides by $(2n -1)(2n + 1)$ gives:

$$1 = A(2n + 1) + B(2n - 1)$$

We can find A and B by substuting $n = ±\frac{1}{2}:$

$$ \begin{align} \text{Let n = }&\frac{1}{2}, \, \text{ then }\; A = \frac{1}{2} \\[5pt] \text{Let n = }-&\frac{1}{2}, \, \text{ then }\; B = -\frac{1}{2} \end{align}$$

So our series can be rewritten as

$$\sum_{n = 1}^{\infty} \frac{1}{(2n - 1)(2n + 1)} = \sum_{n = 1}^{\infty} \left( \frac{1}{4n - 2} - \frac{1}{4n + 2} \right)$$

Now let's expand the sum and take the limit as the number of terms → ∞

$$ \begin{align} \lim_{N\to\infty} \sum_{n = 1}^N &\left( \frac{1}{4n - 2} - \frac{1}{4n + 2} \right) \\[5pt] &= \lim_{N\to\infty}\left( \frac{1}{2} - \frac{1}{6} \right) + \left( \frac{1}{6} - \frac{1}{10} \right) + \left( \frac{1}{10} - \frac{1}{14} \right) + \dots + \left( \frac{1}{4N - 2} - \frac{1}{4N + 2} \right) \\[5pt] &= \lim_{N\to\infty}\frac{1}{2} + \left( \frac{1}{6} - \frac{1}{6} \right) + \left( \frac{1}{10} - \frac{1}{10} \right) + \left( \frac{1}{14} - \frac{1}{14} \right) + \dots - \frac{1}{4N + 2} \\[5pt] &= \lim_{N\to\infty} \left[ \frac{1}{2} - \frac{1}{4N + 2}\right] = \bf \frac{1}{2} \end{align}$$

Example 2

Find the sum of the series $\sum_{n = 1}^{\infty} \frac{1}{n^2 + 7n + 12}.$

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Solution: The denominator is factorable, which allows us to decompose this rational function into the sum (hopefully difference) of two:

$$ \begin{align} \frac{1}{n^2 + 7n + 12} &= \frac{1}{(n + 3)(n + 4)} \\[3pt] \frac{1}{(n + 3)(n + 4)} &= \frac{A}{n + 3} + \frac{B}{n + 4} \\[3pt] 1 &= A(n + 4) + B(n + 3) \\[3pt] \text{Let n = -4, then} \; B &= -1 \\[3pt] \text{Let n = -3, then} \; A &= 1 \end{align}$$

Now we can expand the series as $n \to\infty :$

$$ \begin{align} \lim_{N\to\infty} &\sum_{n = 1}^N \left( \frac{1}{n + 3} - \frac{1}{n + 4} \right) \\[5pt] &= \lim_{N\to\infty} \left( \frac{1}{4} - \frac{1}{5} \right) + \left( \frac{1}{5} - \frac{1}{6} \right) + \left( \frac{1}{6} - \frac{1}{7} \right) + \dots + \left( \frac{1}{n + 3} - \frac{1}{n + 4} \right) \\[5pt] &= \lim_{N\to\infty} \frac{1}{4} + \left( \frac{1}{5} - \frac{1}{5} \right) + \left( \frac{1}{6} - \frac{1}{6} \right) + \left( \frac{1}{7} - \frac{1}{7} \right) + \dots - \frac{1}{n + 4} \\[5pt] &= \lim_{N\to\infty} = \left( \frac{1}{4} - \frac{1}{n + 4} \right) = \bf \frac{1}{4} \end{align}$$

Example 3

Calculate the sum of the first 100 terms of $\sum \frac{1}{n^2 + 3n + 2}.$

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Solution: First we can factor the denominator, which will allow us to decompose this rational function into two:

$$\frac{1}{n^2 + 3n + 2} = \frac{1}{(n + 1)(n + 2)}$$

Now let's decompose:

$$ \begin{align} \frac{1}{(n + 1)(n + 2)} &= \frac{A}{n + 1} + \frac{B}{n + 2}\\[5pt] 1 &= A(n + 2) + B(n + 1)\\[5pt] \text{Let }\; n &= -1, \; \text{ then } \; A = 1 \\[5pt] \text{Let }\; n &= -2, \; \text{ then } \; B = -1, \; \text{ so} \\[5pt] \frac{1}{(n + 1)(n + 2)} &= \frac{1}{n + 1} - \frac{1}{n + 2} \end{align}$$

Now we'll telescope the series:

$$ \begin{align} S &= \left( \frac{1}{2} - \frac{1}{3} \right) + \left( \frac{1}{3} - \frac{1}{4} \right) + \left( \frac{1}{4} - \frac{1}{5} \right) + \dots + \left( \frac{1}{101} - \frac{1}{102} \right)\\[5pt] &= \frac{1}{2} + \left( \frac{1}{3} - \frac{1}{3} \right) + \left( \frac{1}{4} - \frac{1}{4} \right) + \left( \frac{1}{5} - \frac{1}{5} \right) + \dots - \frac{1}{102}\\[5pt] &= \frac{1}{2} - \frac{1}{102} = \frac{51 - 1}{102} = \frac{25}{51} \approx \frac{1}{2} \end{align}$$

It's easy to see that the sum of this series as n → ∞ is ½.