This is a beta course, so its structure, chapters, and examples may continue to change.
The Telescoping Method
The telescoping method is one of the most ingenious methods for summing sequences. Its core idea is to “split” each term into the difference of two terms, so that most terms cancel each other during the summation, leaving only the first and last few terms.
The Principle of the Method Split the general term a n a_n a n a n = f ( n ) − f ( n + k ) a_n = f(n) - f(n+k) a n = f ( n ) − f ( n + k )
Basic idea :
S n = a 1 + a 2 + a 3 + ⋯ + a n = [ f ( 1 ) − f ( 2 ) ] + [ f ( 2 ) − f ( 3 ) ] + [ f ( 3 ) − f ( 4 ) ] + ⋯ + [ f ( n ) − f ( n + 1 ) ] = f ( 1 ) − f ( n + 1 ) \begin{aligned}
S_n &= a_1 + a_2 + a_3 + \cdots + a_n \\
&= [f(1) - f(2)] + [f(2) - f(3)] + [f(3) - f(4)] + \cdots + [f(n) - f(n+1)] \\
&= f(1) - f(n+1)
\end{aligned} S n = a 1 + a 2 + a 3 + ⋯ + a n = [ f ( 1 ) − f ( 2 )] + [ f ( 2 ) − f ( 3 )] + [ f ( 3 ) − f ( 4 )] + ⋯ + [ f ( n ) − f ( n + 1 )] = f ( 1 ) − f ( n + 1 ) The intermediate f ( 2 ) , f ( 3 ) , … , f ( n ) f(2), f(3), \ldots, f(n) f ( 2 ) , f ( 3 ) , … , f ( n )
Generalization :
1 n ( n + k ) = 1 k ( 1 n − 1 n + k ) \frac{1}{n(n+k)} = \frac{1}{k}\left(\frac{1}{n} - \frac{1}{n+k}\right) n ( n + k ) 1 = k 1 ( n 1 − n + k 1 )
1 n + n + 1 = n + 1 − n \frac{1}{\sqrt{n} + \sqrt{n+1}} = \sqrt{n+1} - \sqrt{n} n + n + 1 1 = n + 1 − n
Derivation : multiply both the numerator and the denominator by n + 1 − n \sqrt{n+1} - \sqrt{n} n + 1 − n
1 ( 2 n − 1 ) ( 2 n + 1 ) = 1 2 ( 1 2 n − 1 − 1 2 n + 1 ) \frac{1}{(2n-1)(2n+1)} = \frac{1}{2}\left(\frac{1}{2n-1} - \frac{1}{2n+1}\right) ( 2 n − 1 ) ( 2 n + 1 ) 1 = 2 1 ( 2 n − 1 1 − 2 n + 1 1 )
How do we determine whether a sequence can be telescoped? Observe the characteristics of the general term:
Fractional form : the denominator is the product of two or more factorsFactor relationship : there is a regular pattern between the factors of the denominator (e.g., they differ by a constant)Try to split : see whether it can be written in the form f ( n ) − f ( n + k ) f(n) - f(n+k) f ( n ) − f ( n + k ) Technique : for 1 n ( n + k ) \frac{1}{n(n+k)} n ( n + k ) 1 1 n ( n + k ) = A n + B n + k \frac{1}{n(n+k)} = \frac{A}{n} + \frac{B}{n+k} n ( n + k ) 1 = n A + n + k B
After putting over a common denominator, compare the coefficients to find A A A B B B
Worked Examples Example 1: Basic Telescoping
Find the sum: S n = 1 1 × 2 + 1 2 × 3 + 1 3 × 4 + ⋯ + 1 n ( n + 1 ) S_n = \frac{1}{1 \times 2} + \frac{1}{2 \times 3} + \frac{1}{3 \times 4} + \cdots + \frac{1}{n(n+1)} S n = 1 × 2 1 + 2 × 3 1 + 3 × 4 1 + ⋯ + n ( n + 1 ) 1
Solution :
Use the telescoping formula 1 n ( n + 1 ) = 1 n − 1 n + 1 \frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1} n ( n + 1 ) 1 = n 1 − n + 1 1
S n = ( 1 1 − 1 2 ) + ( 1 2 − 1 3 ) + ( 1 3 − 1 4 ) + ⋯ + ( 1 n − 1 n + 1 ) = 1 1 − 1 n + 1 = n n + 1 \begin{aligned}
S_n &= \left(\frac{1}{1} - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{4}\right) + \cdots + \left(\frac{1}{n} - \frac{1}{n+1}\right) \\
&= \frac{1}{1} - \frac{1}{n+1} \\
&= \frac{n}{n+1}
\end{aligned} S n = ( 1 1 − 2 1 ) + ( 2 1 − 3 1 ) + ( 3 1 − 4 1 ) + ⋯ + ( n 1 − n + 1 1 ) = 1 1 − n + 1 1 = n + 1 n Example 2: Telescoping with a Coefficient
Find the sum: S n = 1 1 × 3 + 1 3 × 5 + 1 5 × 7 + ⋯ + 1 ( 2 n − 1 ) ( 2 n + 1 ) S_n = \frac{1}{1 \times 3} + \frac{1}{3 \times 5} + \frac{1}{5 \times 7} + \cdots + \frac{1}{(2n-1)(2n+1)} S n = 1 × 3 1 + 3 × 5 1 + 5 × 7 1 + ⋯ + ( 2 n − 1 ) ( 2 n + 1 ) 1
Solution :
Use the formula 1 ( 2 n − 1 ) ( 2 n + 1 ) = 1 2 ( 1 2 n − 1 − 1 2 n + 1 ) \frac{1}{(2n-1)(2n+1)} = \frac{1}{2}\left(\frac{1}{2n-1} - \frac{1}{2n+1}\right) ( 2 n − 1 ) ( 2 n + 1 ) 1 = 2 1 ( 2 n − 1 1 − 2 n + 1 1 )
S n = 1 2 [ ( 1 1 − 1 3 ) + ( 1 3 − 1 5 ) + ⋯ + ( 1 2 n − 1 − 1 2 n + 1 ) ] = 1 2 ( 1 1 − 1 2 n + 1 ) = 1 2 ⋅ 2 n 2 n + 1 = n 2 n + 1 \begin{aligned}
S_n &= \frac{1}{2}\left[\left(\frac{1}{1} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{5}\right) + \cdots + \left(\frac{1}{2n-1} - \frac{1}{2n+1}\right)\right] \\
&= \frac{1}{2}\left(\frac{1}{1} - \frac{1}{2n+1}\right) \\
&= \frac{1}{2} \cdot \frac{2n}{2n+1} \\
&= \frac{n}{2n+1}
\end{aligned} S n = 2 1 [ ( 1 1 − 3 1 ) + ( 3 1 − 5 1 ) + ⋯ + ( 2 n − 1 1 − 2 n + 1 1 ) ] = 2 1 ( 1 1 − 2 n + 1 1 ) = 2 1 ⋅ 2 n + 1 2 n = 2 n + 1 n Example 3: Radical Telescoping
Find the sum: S n = 1 1 + 2 + 1 2 + 3 + ⋯ + 1 n + n + 1 S_n = \frac{1}{\sqrt{1} + \sqrt{2}} + \frac{1}{\sqrt{2} + \sqrt{3}} + \cdots + \frac{1}{\sqrt{n} + \sqrt{n+1}} S n = 1 + 2 1 + 2 + 3 1 + ⋯ + n + n + 1 1
Solution :
Multiply both the numerator and the denominator by n + 1 − n \sqrt{n+1} - \sqrt{n} n + 1 − n
1 n + n + 1 = n + 1 − n ( n + 1 + n ) ( n + 1 − n ) = n + 1 − n \frac{1}{\sqrt{n} + \sqrt{n+1}} = \frac{\sqrt{n+1} - \sqrt{n}}{(\sqrt{n+1} + \sqrt{n})(\sqrt{n+1} - \sqrt{n})} = \sqrt{n+1} - \sqrt{n} n + n + 1 1 = ( n + 1 + n ) ( n + 1 − n ) n + 1 − n = n + 1 − n
Therefore:
S n = ( 2 − 1 ) + ( 3 − 2 ) + ⋯ + ( n + 1 − n ) = n + 1 − 1 \begin{aligned}
S_n &= (\sqrt{2} - \sqrt{1}) + (\sqrt{3} - \sqrt{2}) + \cdots + (\sqrt{n+1} - \sqrt{n}) \\
&= \sqrt{n+1} - 1
\end{aligned} S n = ( 2 − 1 ) + ( 3 − 2 ) + ⋯ + ( n + 1 − n ) = n + 1 − 1 Practice Problems Exercise 1
Find the sum: S n = 1 2 × 4 + 1 4 × 6 + 1 6 × 8 + ⋯ + 1 2 n ( 2 n + 2 ) S_n = \frac{1}{2 \times 4} + \frac{1}{4 \times 6} + \frac{1}{6 \times 8} + \cdots + \frac{1}{2n(2n+2)} S n = 2 × 4 1 + 4 × 6 1 + 6 × 8 1 + ⋯ + 2 n ( 2 n + 2 ) 1
Reference Answer sequence summation telescoping
Approach : first extract the common factor, then telescope.
Steps :
1 2 n ( 2 n + 2 ) = 1 2 n ⋅ 2 ( n + 1 ) = 1 4 n ( n + 1 ) = 1 4 ( 1 n − 1 n + 1 ) \frac{1}{2n(2n+2)} = \frac{1}{2n \cdot 2(n+1)} = \frac{1}{4n(n+1)} = \frac{1}{4}\left(\frac{1}{n} - \frac{1}{n+1}\right) 2 n ( 2 n + 2 ) 1 = 2 n ⋅ 2 ( n + 1 ) 1 = 4 n ( n + 1 ) 1 = 4 1 ( n 1 − n + 1 1 )
S n = 1 4 [ ( 1 1 − 1 2 ) + ( 1 2 − 1 3 ) + ⋯ + ( 1 n − 1 n + 1 ) ] = 1 4 ( 1 − 1 n + 1 ) = n 4 ( n + 1 ) \begin{aligned}
S_n &= \frac{1}{4}\left[\left(\frac{1}{1} - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \cdots + \left(\frac{1}{n} - \frac{1}{n+1}\right)\right] \\
&= \frac{1}{4}\left(1 - \frac{1}{n+1}\right) \\
&= \frac{n}{4(n+1)}
\end{aligned} S n = 4 1 [ ( 1 1 − 2 1 ) + ( 2 1 − 3 1 ) + ⋯ + ( n 1 − n + 1 1 ) ] = 4 1 ( 1 − n + 1 1 ) = 4 ( n + 1 ) n Answer : S n = n 4 ( n + 1 ) S_n = \frac{n}{4(n+1)} S n = 4 ( n + 1 ) n
Exercise 2
Find the sum: S n = 1 1 × 4 + 1 4 × 7 + 1 7 × 10 + ⋯ + 1 ( 3 n − 2 ) ( 3 n + 1 ) S_n = \frac{1}{1 \times 4} + \frac{1}{4 \times 7} + \frac{1}{7 \times 10} + \cdots + \frac{1}{(3n-2)(3n+1)} S n = 1 × 4 1 + 4 × 7 1 + 7 × 10 1 + ⋯ + ( 3 n − 2 ) ( 3 n + 1 ) 1
Reference Answer sequence summation telescoping
Approach : the denominators differ by 3, so use the formula 1 n ( n + k ) = 1 k ( 1 n − 1 n + k ) \frac{1}{n(n+k)} = \frac{1}{k}(\frac{1}{n} - \frac{1}{n+k}) n ( n + k ) 1 = k 1 ( n 1 − n + k 1 )
Steps :
1 ( 3 n − 2 ) ( 3 n + 1 ) = 1 3 ( 1 3 n − 2 − 1 3 n + 1 ) \frac{1}{(3n-2)(3n+1)} = \frac{1}{3}\left(\frac{1}{3n-2} - \frac{1}{3n+1}\right) ( 3 n − 2 ) ( 3 n + 1 ) 1 = 3 1 ( 3 n − 2 1 − 3 n + 1 1 )
S n = 1 3 [ ( 1 1 − 1 4 ) + ( 1 4 − 1 7 ) + ⋯ + ( 1 3 n − 2 − 1 3 n + 1 ) ] = 1 3 ( 1 − 1 3 n + 1 ) = n 3 n + 1 \begin{aligned}
S_n &= \frac{1}{3}\left[\left(\frac{1}{1} - \frac{1}{4}\right) + \left(\frac{1}{4} - \frac{1}{7}\right) + \cdots + \left(\frac{1}{3n-2} - \frac{1}{3n+1}\right)\right] \\
&= \frac{1}{3}\left(1 - \frac{1}{3n+1}\right) \\
&= \frac{n}{3n+1}
\end{aligned} S n = 3 1 [ ( 1 1 − 4 1 ) + ( 4 1 − 7 1 ) + ⋯ + ( 3 n − 2 1 − 3 n + 1 1 ) ] = 3 1 ( 1 − 3 n + 1 1 ) = 3 n + 1 n Answer : S n = n 3 n + 1 S_n = \frac{n}{3n+1} S n = 3 n + 1 n
Exercise 3
Find the sum: S n = 1 1 + 3 + 1 3 + 5 + ⋯ + 1 2 n − 1 + 2 n + 1 S_n = \frac{1}{\sqrt{1} + \sqrt{3}} + \frac{1}{\sqrt{3} + \sqrt{5}} + \cdots + \frac{1}{\sqrt{2n-1} + \sqrt{2n+1}} S n = 1 + 3 1 + 3 + 5 1 + ⋯ + 2 n − 1 + 2 n + 1 1
Reference Answer sequence summation telescoping
Approach : multiply both the numerator and the denominator by the conjugate.
Steps :
1 2 n − 1 + 2 n + 1 = 2 n + 1 − 2 n − 1 2 \frac{1}{\sqrt{2n-1} + \sqrt{2n+1}} = \frac{\sqrt{2n+1} - \sqrt{2n-1}}{2} 2 n − 1 + 2 n + 1 1 = 2 2 n + 1 − 2 n − 1
S n = 1 2 [ ( 3 − 1 ) + ( 5 − 3 ) + ⋯ + ( 2 n + 1 − 2 n − 1 ) ] = 1 2 ( 2 n + 1 − 1 ) \begin{aligned}
S_n &= \frac{1}{2}[(\sqrt{3} - \sqrt{1}) + (\sqrt{5} - \sqrt{3}) + \cdots + (\sqrt{2n+1} - \sqrt{2n-1})] \\
&= \frac{1}{2}(\sqrt{2n+1} - 1)
\end{aligned} S n = 2 1 [( 3 − 1 ) + ( 5 − 3 ) + ⋯ + ( 2 n + 1 − 2 n − 1 )] = 2 1 ( 2 n + 1 − 1 ) Answer : S n = 2 n + 1 − 1 2 S_n = \frac{\sqrt{2n+1} - 1}{2} S n = 2 2 n + 1 − 1
Exercise 4
Find the sum: S n = 1 1 × 2 × 3 + 1 2 × 3 × 4 + ⋯ + 1 n ( n + 1 ) ( n + 2 ) S_n = \frac{1}{1 \times 2 \times 3} + \frac{1}{2 \times 3 \times 4} + \cdots + \frac{1}{n(n+1)(n+2)} S n = 1 × 2 × 3 1 + 2 × 3 × 4 1 + ⋯ + n ( n + 1 ) ( n + 2 ) 1
Reference Answer sequence summation telescoping
Approach : for a product of three consecutive integers, telescoping needs to be done twice.
Steps :
1 n ( n + 1 ) ( n + 2 ) = 1 2 [ 1 n ( n + 1 ) − 1 ( n + 1 ) ( n + 2 ) ] \frac{1}{n(n+1)(n+2)} = \frac{1}{2}\left[\frac{1}{n(n+1)} - \frac{1}{(n+1)(n+2)}\right] n ( n + 1 ) ( n + 2 ) 1 = 2 1 [ n ( n + 1 ) 1 − ( n + 1 ) ( n + 2 ) 1 ]
S n = 1 2 [ ( 1 1 × 2 − 1 2 × 3 ) + ( 1 2 × 3 − 1 3 × 4 ) + ⋯ ] = 1 2 ( 1 1 × 2 − 1 ( n + 1 ) ( n + 2 ) ) = 1 2 ( 1 2 − 1 ( n + 1 ) ( n + 2 ) ) = 1 4 − 1 2 ( n + 1 ) ( n + 2 ) = n ( n + 3 ) 4 ( n + 1 ) ( n + 2 ) \begin{aligned}
S_n &= \frac{1}{2}\left[\left(\frac{1}{1 \times 2} - \frac{1}{2 \times 3}\right) + \left(\frac{1}{2 \times 3} - \frac{1}{3 \times 4}\right) + \cdots\right] \\
&= \frac{1}{2}\left(\frac{1}{1 \times 2} - \frac{1}{(n+1)(n+2)}\right) \\
&= \frac{1}{2}\left(\frac{1}{2} - \frac{1}{(n+1)(n+2)}\right) \\
&= \frac{1}{4} - \frac{1}{2(n+1)(n+2)} \\
&= \frac{n(n+3)}{4(n+1)(n+2)}
\end{aligned} S n = 2 1 [ ( 1 × 2 1 − 2 × 3 1 ) + ( 2 × 3 1 − 3 × 4 1 ) + ⋯ ] = 2 1 ( 1 × 2 1 − ( n + 1 ) ( n + 2 ) 1 ) = 2 1 ( 2 1 − ( n + 1 ) ( n + 2 ) 1 ) = 4 1 − 2 ( n + 1 ) ( n + 2 ) 1 = 4 ( n + 1 ) ( n + 2 ) n ( n + 3 ) Answer : S n = n ( n + 3 ) 4 ( n + 1 ) ( n + 2 ) S_n = \frac{n(n+3)}{4(n+1)(n+2)} S n = 4 ( n + 1 ) ( n + 2 ) n ( n + 3 )
Summary Symbols Used in This Article
Symbol Type Pronunciation/Explanation Meaning in This Article S n S_n S n Summation symbol S sub n The sum of the first n n n f ( n ) f(n) f ( n ) Function symbol f of n The function form after telescoping k k k Constant k The telescoping interval
