Advanced book on Mathematics Olympiad

122 3 Real Analysis

1

ak+ 1

=

1

ak− 1

−

1

ak+ 1 − 1

, fork≥ 1.

Summing up these equalities fork= 1 , 2 ,...,nyields

1

a 1 + 1

+···+

1

an+ 1

=

1

a 1 − 1

−

1

a 2 − 1

+

1

a 2 − 1

−

1

a 3 − 1

+···

+

1

an− 1

−

1

an+ 1 − 1

=

1

2

−

1

an+ 1 − 1

.

Finally, adda 0 +^11 +an+^11 − 1 to both sides to obtain the identity from the statement. 

Example.Express

∑^49

n= 1

1

√

n+

√

n^2 − 1

asa+b

√

2 for some integersaandb.

Solution.We have

1

√

n+

√

n^2 − 1

=

1

√(√

n+ 1

2 +

√

n− 1

2

) 2 =

1

√

n+ 1

2 +

√

n− 1

2

=

√

n+ 1

2 −

√

n− 1

2

n+ 1

2 −

n− 1

2

=

√

n+ 1

2

−

√

n− 1

2

.

Hence the sum from the statement telescopes to
√
49 + 1
2

+

√

48 + 1

2

−

√

1

2

− 0 = 5 +

7

√

2

−

1

√

2

= 5 + 3

√

2. 

Apply the telescopic method to the following problems.

365.Prove the identity

∑n

k= 1

(k^2 + 1 )k!=n(n+ 1 )!.

366.Letζbe a root of unity. Prove that

ζ−^1 =

∑∞

n= 0

ζn( 1 −ζ)( 1 −ζ^2 )···( 1 −ζn),

with the convention that the 0th term of the series is 1.