Here is my solution

QUESTION 14110>
Since lim sup
n 

an
  and
bn



 bn  M 2 . Thus,
n 1



b
n 1





n 1

n 1

 an  

n

  . There must be M 1 , M 2 



such that lim sup

an
a 
a 
bn  sup n   bn  lim sup n   bn  M1M 2 which is a finite
n 
bn
bn n 1
bn n 1



number. Hence,

a
n 1

n

converges.

QUESTION 14113>
1. The tangent vector can be computed as follow

r  t   2sin  2t  i  2cos  2t  j
To get the unit tangent vector we need the length of the tangent vector

 2sin  2t    2cos  2t 

r  t  

2

2

 2 sin 2  2t   cos2  2t   2

The unit tangent vector is then,

T t  

r   t  2sin  2t  i  2cos  2t  j

  sin  2t  i  cos  2t  j
2
r t 

2. We differentiate the unit tangent vector found above

N  t   T   t   2cos  2t  i  2sin  2t  j
We need the length T   t  to get the principal normal vector

T  t  

 2cos  2t    2sin  2t 
2

2

 2 cos2  2t   sin 2  2t   2

Hence, the principal normal vector is given by

N t  

T   t  2cos  2t  i  2sin  2t  j

  cos  2t  i  sin  2t  j
2
T  t 

3. Hence, the curvature can be computed using the formula



T  t 
r t 



n 

an
 M1 and
bn

2
 1 , which is constant function.
2

QUESTION 14114>

Let S a be the area of the lune that we need to find, and let S b be the area of the circular segment
formed by the diameter of the blue circle treated as a chord of the red circle.

Sa 

 r2
2

 Sb

Sb  Ssec tor  S 
1
Ssec tor   R 2
2
1 2
r
r
 , so Ssec tor  R arcsin  
2
R
R

We have   2 arcsin 

1
2r R 2  r 2  r R 2  r 2
2
r
 Sb  Ssec tor  S   R 2 arcsin    r R 2  r 2
R
2
2
r
r  2
 r2
r
r
2
2 
 Sa 
 Sb 
  R arcsin    r R  r  
 r R 2  r 2  R 2 arcsin  
2
2 
2
R
R

S 

Hence, the area of the crescent-shaped region (called a lune) bounded by arcs of circles with radii r and
R with r  R equal to Sa 

 r2

r
 r R 2  r 2  R 2 arcsin ...