Real Analysis 499=
1
4
(√
n^2 +(n+ 1 )^2 −√
n^2 +(n− 1 )^2)
=
1
4
(bn+ 1 −bn),wherebn=
√
n^2 +(n− 1 )^2. Hence the given sum collapses to^14 ( 29 − 1 )= 7.
(Mathematical Reflections, proposed by T. Andreescu)372.Let us look at the summation overnfirst. Multiplying each term by(m+n+ 2 )−
(n+ 1 )and dividing bym+1, we obtain
m!
m+ 1∑∞
n= 0(
n!
(m+n+ 1 )!−
(n+ 1 )!
(m+n+ 2 )!)
.
This is a telescopic sum that adds up to
m!
m+ 1·
0!
(m+ 1 )!.
Consequently, the expression we are computing is equal to
∑∞m= 01
(m+ 1 )^2=
π^2
6.
(Mathematical Mayhem, 1995)373.This problem is similar to the last example from the introduction. We start with
ak−bk=1
2
[
4 k+(k+ 1 )+(k− 1 )− 4√
k^2 +k+ 4√
k^2 −k+ 2√
k^2 − 1]
=
1
2
(
2
√
k−√
k+ 1 −√
k− 1) 2
.
From here we obtain
√
ak−bk=1
√
2
(
2
√
k−√
k+ 1 −√
k− 1)
=−
1
√
2
(√
k+ 1 −√
k)
+
1
√
2
(√
k−√
k+ 1)
.
The sum from the statement telescopes to
−
1
√
2
(√
50 −
√
1
)
+
1
√
2
(√
49 −
√
0
)
=− 5 + 4
√
2.
(Romanian Mathematical Olympiad, 2004, proposed by T. Andreescu)