From Wikipedia, the free encyclopedia
In mathematics , telescoping series is an informal expression referring to a series whose sum can be found by exploiting the circumstance that nearly every term cancels with a succeeding or preceding term. Such a technique is also known as the method of differences .
For example, the series
∑
n
=
1
∞
1
n
(
n
+
1
)
{\displaystyle \sum _{n=1}^{\infty }{\frac {1}{n(n+1)}}}
simplifies as
∑
n
=
1
∞
1
n
(
n
+
1
)
=
∑
n
=
1
∞
(
1
n
−
1
n
+
1
)
{\displaystyle \sum _{n=1}^{\infty }{\frac {1}{n(n+1)}}=\sum _{n=1}^{\infty }\left({\frac {1}{n}}-{\frac {1}{n+1}}\right)\,}
=
(
1
−
1
2
)
+
(
1
2
−
1
3
)
+
⋯
{\displaystyle =\left(1-{\frac {1}{2}}\right)+\left({\frac {1}{2}}-{\frac {1}{3}}\right)+\cdots \,}
=
1
+
(
−
1
2
+
1
2
)
+
(
−
1
3
+
1
3
)
+
⋯
=
1.
{\displaystyle =1+\left(-{\frac {1}{2}}+{\frac {1}{2}}\right)+\left(-{\frac {1}{3}}+{\frac {1}{3}}\right)+\cdots =1.\,}
A pitfall While telescoping is a neat technique, there are pitfalls to watch out for:
0
=
∑
n
=
1
∞
0
=
∑
n
=
1
∞
(
1
−
1
)
=
1
+
∑
n
=
1
∞
(
−
1
+
1
)
=
1
{\displaystyle 0=\sum _{n=1}^{\infty }0=\sum _{n=1}^{\infty }(1-1)=1+\sum _{n=1}^{\infty }(-1+1)=1\,}
is not correct because regrouping of terms is invalid unless the individual terms converge to 0. The way to avoid this error is to find the sum of the first N terms first and then take the limit as N approaches infinity:
∑
n
=
1
N
1
n
(
n
+
1
)
=
∑
n
=
1
N
(
1
n
−
1
n
+
1
)
{\displaystyle \sum _{n=1}^{N}{\frac {1}{n(n+1)}}=\sum _{n=1}^{N}\left({\frac {1}{n}}-{\frac {1}{n+1}}\right)\,}
=
(
1
−
1
2
)
+
(
1
2
−
1
3
)
+
⋯
+
(
1
N
−
1
N
+
1
)
{\displaystyle =\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
+
(
−
1
2
+
1
2
)
+
(
−
1
3
+
1
3
)
+
⋯
+
(
−
1
N
+
1
N
)
−
1
N
+
1
{\displaystyle =1+\left(-{\frac {1}{2}}+{\frac {1}{2}}\right)+\left(-{\frac {1}{3}}+{\frac {1}{3}}\right)+\cdots +\left(-{\frac {1}{N}}+{\frac {1}{N}}\right)-{\frac {1}{N+1}}\,}
=
1
−
1
N
+
1
→
1
a
s
N
→
∞
.
{\displaystyle =1-{\frac {1}{N+1}}\to 1\ \mathrm {as} \ N\to \infty .\,}
More examples Many trigonometric functions also admit representation as a difference, which allows telescopic cancelling between the consequent terms.
∑
n
=
1
N
sin
(
n
)
=
∑
n
=
1
N
1
2
csc
(
1
2
)
(
2
sin
(
1
2
)
sin
(
n
)
)
{\displaystyle \sum _{n=1}^{N}\sin \left(n\right)=\sum _{n=1}^{N}{\frac {1}{2}}\csc \left({\frac {1}{2}}\right)\left(2\sin \left({\frac {1}{2}}\right)\sin \left(n\right)\right)}
=
1
2
csc
(
1
2
)
∑
n
=
1
N
(
cos
(
2
n
−
1
2
)
−
cos
(
2
n
+
1
2
)
)
{\displaystyle ={\frac {1}{2}}\csc \left({\frac {1}{2}}\right)\sum _{n=1}^{N}\left(\cos \left({\frac {2n-1}{2}}\right)-\cos \left({\frac {2n+1}{2}}\right)\right)}
=
1
2
csc
(
1
2
)
(
cos
(
1
2
)
−
cos
(
2
N
+
1
2
)
)
.
{\displaystyle ={\frac {1}{2}}\csc \left({\frac {1}{2}}\right)\left(\cos \left({\frac {1}{2}}\right)-\cos \left({\frac {2N+1}{2}}\right)\right).}
∑
n
=
1
N
f
(
n
)
g
(
n
)
,
{\displaystyle \sum _{n=1}^{N}{f(n) \over g(n)},}
where f and g are polynomial functions whose quotient may be broken up into partial fractions , will fail to admit summation by this method. In particular, we have
∑
n
=
0
∞
2
n
+
3
(
n
+
1
)
(
n
+
2
)
{\displaystyle \sum _{n=0}^{\infty }{\frac {2n+3}{(n+1)(n+2)}}}
=
∑
n
=
0
∞
(
1
n
+
1
+
1
n
+
2
)
{\displaystyle =\sum _{n=0}^{\infty }\left({\frac {1}{n+1}}+{\frac {1}{n+2}}\right)}
=
1
1
+
1
2
+
1
2
+
1
3
+
1
3
+
1
4
+
⋯
+
1
n
−
1
+
1
n
+
1
n
+
1
n
+
1
+
1
n
+
1
+
1
n
+
2
+
⋯
{\displaystyle ={\frac {1}{1}}+{\frac {1}{2}}+{\frac {1}{2}}+{\frac {1}{3}}+{\frac {1}{3}}+{\frac {1}{4}}+\cdots +{\frac {1}{n-1}}+{\frac {1}{n}}+{\frac {1}{n}}+{\frac {1}{n+1}}+{\frac {1}{n+1}}+{\frac {1}{n+2}}+\cdots }
=
∞
.
{\displaystyle =\infty .}
The problem is that the terms do not cancel. Let
k
{\displaystyle k}
∑
n
=
1
∞
1
n
(
n
+
k
)
=
H
k
k
{\displaystyle \sum _{n=1}^{\infty }{\frac {1}{n(n+k)}}={\frac {H_{k}}{k}}}
where
H
k
{\displaystyle H_{k}}
harmonic number . 
