Advanced book on Mathematics Olympiad

3.1 Sequences and Series 121

∫b

a

f(t)dt=F(b)−F(a), whereF′(t)=f(t),

becomes the telescopic method for summing a series

∑n

1

ak=bn+ 1 −b 1 , whereak=bk+ 1 −bk,k≥ 1.

As in the case of integrals, when applying the telescopic method to a series, the struggle is
to find the “antiderivative’’ of the general term. But compared to the case of integrals, here
we lack an algorithmic way. This is what makes such problems attractive for mathematics
competitions. A simple example that comes to mind is the following.

Example.Find the sum

1

√

1 +

√

2

+

1

√

2 +

√

3

+···+

1

√

n+

√

n+ 1

.

Solution.The “antiderivative’’ of the general term of the sum is found by rationalizing

the denominator:

1

√

k+

√

k+ 1

=

√

k+ 1 −

√

k

k+ 1 −k

=

√

k+ 1 −

√

k.

The sum is therefore equal to

(

√

2 −

√

1 )+(

√

3 −

√

2 )+···+(

√

n+ 1 −

√

n)=

√

n+ 1 − 1. 

Not all problems are so simple, as the next two examples show.

Example.Leta 0 =1,a 1 =3,an+ 1 =a

(^2) n+ 1
2 ,n≥1. Prove that
1
a 0 + 1

+

1

a 1 + 1

+···+

1

an+ 1

+

1

an+ 1 − 1

= 1 , for alln≥ 1.

Solution.We have

ak+ 1 − 1 =

a^2 k− 1

2

,

so

1

ak+ 1 − 1

=

1

ak− 1

−

1

ak+ 1

, fork≥ 1.

This allows us to express the terms of the sum from the statement as “derivatives’’: