112 3 Real Analysis
Next, we present a famous identity of S.A. Ramanujan.Example.Prove that
√
1 + 2√
1 + 3
√
1 + 4
√
1 +··· = 3.
Solution.We approach the problem in more generality by introducing the functionf:
[ 1 ,∞)→R,
f(x)=√
1 +x√
1 +(x+ 1 )√
1 +(x+ 2 )√
1 +···.
Is this function well defined? Truncating tonsquare roots, we obtain an increasing
sequence. All we need to show is that this sequence is bounded from above. And it is,
because
f(x)≤√
(x+ 1 )√
(x+ 2 )√
(x+ 3 )···≤
√
2 x√
3 x√
4 x··· ≤√
2 x√
4 x√
8 x···= 2∑k
2 kx∑ 1
2 k≤ 2(^12) + (^12) + (^14) + (^14) + (^18) + (^18) +···
x= 2 x.
This shows, moreover, thatf(x)≤ 2 x, forx≥1. Note also that
f(x)≥
√
x√
x√
x··· =x.For reasons that will become apparent, we weaken this inequality tof(x)≥^12 (x+ 1 ).
We then square the defining relation and obtain the functional equation
(f (x))^2 =xf (x+ 1 )+ 1.Combining this with
1
2
(x+ 1 )≤f(x+ 1 )≤ 2 (x+ 1 ),we obtain
x·
x+ 1
2+ 1 ≤(f (x))^2 ≤ 2 x(x+ 1 )+ 1 ,which yields the sharper double inequality