Advanced book on Mathematics Olympiad

500 Real Analysis

374.First solution: LetSn=

∑n

k= 0 (−^1 )

k(n−k)!(n+k)!.Reordering the terms of the

sum, we have

Sn=(− 1 )n

∑n

k= 0

(− 1 )kk!( 2 n−k)!

=(− 1 )n

1

2

(

(− 1 )nn!n!+

∑^2 n

k= 0

(− 1 )kk!( 2 n−k)!

)

=

(n!)^2

2

+(− 1 )n

Tn

2

,

whereTn=

∑ 2 n

k= 0 (−^1 )

kk!( 2 n−k)!. We now focus on the sumTn. Observe that

Tn

( 2 n)!

=

∑^2 n

k= 0

(− 1 )k

( 2 n

k

)

and

1

( 2 n

k

)=

2 n+ 1

2 (n+ 1 )

[

1

( 2 n+ 1

k

)+

1

( 2 n+ 1

k+ 1

)

]

.

Hence

Tn

( 2 n)!

=

2 n+ 1

2 (n+ 1 )

[

1

( 2 n+ 1

0

)+

1

( 2 n+ 1

1

)−

1

( 2 n+ 1

1

)−

1

( 2 n+ 1

2

)+···+

1

( 2 n+ 1

2 n

)+

1

( 2 n+ 1

2 n+ 1

)

]

.

This sum telescopes to

2 n+ 1

2 (n+ 1 )

[

1

( 2 n+ 1

0

)+

1

( 2 n+ 1

2 n+ 1

)

]

=

2 n+ 1

n+ 1

.

ThusTn=(^2 nn++ 11 )!, and therefore

Sn=

(n!)^2

2

+(− 1 )n

( 2 n+ 1 )!

2 (n+ 1 )

.

Second solution: Multiply thekth term inSnby(n−k+ 1 )+(n+k+ 1 )and divide by
2 (n+ 1 )to obtain

Sn=

1

2 (n+ 1 )

∑n

k= 0

[

(− 1 )k(n−k+ 1 )!(n+k)!+(− 1 )k(n−k)!(n+k+ 1 )!

]

.