Geometry and Trigonometry 669= lim
N→∞
(arctan(N+ 1 )+arctanN−arctan 1−arctan 0)=
π
2+
π
2−
π
4=
3 π
4.
The sum in part (b) is only slightly more complicated. In the above-mentioned formula
for the difference of arctangents we have to substitutex=(n√+ 21 )^2 andy=(n√− 21 )^2. This
is because
8 n
n^4 − 2 n^2 + 5=
8 n
4 +(n^2 − 1 )^2=
2 [(n+ 1 )^2 −(n− 1 )^2 ]
4 −(n+ 1 )^2 (n− 1 )^2=
(
n√+ 1
2) 2
−
(
n√− 1
2) 2
1 −
(
n√+ 1
2) 2 (
n√− 1
2) 2.
The sum telescopes as
∑∞n= 1arctan8 n
n^4 − 2 n^2 + 5= lim
N→∞∑N
n= 1arctan
8 n
n^4 − 2 n^2 + 5= lim
N→∞∑N
n= 1[
arctan(
n+ 1
√
2) 2
−arctan(
n− 1
√
2) 2 ]
= lim
N→∞[
arctan(
N+ 1
√
2
) 2
+arctan(
N
√
2
) 2
−arctan 0−arctan1
2
]
=π−arctan1
2
.
(American Mathematical Monthly, proposed by J. Anglesio)692.In order for the series to telescope, we wish to write the general term in the form
arcsinbn−arcsinbn+ 1. To determinebnlet us apply the sine function and write
√
n+ 1 −√
n
√
n+ 2√
n+ 1=sinun=bn√
1 −b^2 n+ 1 −bn+ 1√
1 −b^2 n.If we choosebn=√n^1 + 1 , then this equality is satisfied. Therefore,
S= lim
N→∞∑N
n= 0(
arcsin1
√
n+ 1−arcsin1
√
n+ 2)
=arcsin 1− lim
N→∞
arcsin1
√
N+ 2
=
π
2.
(TheMathematics GazetteCompetition, Bucharest, 1927)693.The radii of the circles satisfy the recurrence relationR 1 =1,Rn+ 1 =Rncos 2 nπ+ 1.
Hence