YES

Problem:
 terms(N) -> cons(recip(sqr(N)))
 sqr(0()) -> 0()
 sqr(s()) -> s()
 dbl(0()) -> 0()
 dbl(s()) -> s()
 add(0(),X) -> X
 add(s(),Y) -> s()
 first(0(),X) -> nil()
 first(s(),cons(Y)) -> cons(Y)

Proof:
 Matrix Interpretation Processor: dim=3
  
  interpretation:
           [0]
   [nil] = [0]
           [0],
   
                     [1 0 1]       
   [first](x0, x1) = [0 0 1]x0 + x1
                     [0 0 0]       ,
   
                   [1 1 0]     [1 1 1]     [0]
   [add](x0, x1) = [0 0 0]x0 + [1 1 1]x1 + [1]
                   [1 1 0]     [1 1 1]     [0],
   
               [1 0 0]     [0]
   [dbl](x0) = [0 0 1]x0 + [0]
               [0 0 0]     [1],
   
         [0]
   [s] = [1]
         [1],
   
         [1]
   [0] = [0]
         [0],
   
                [1 1 0]     [0]
   [cons](x0) = [0 1 0]x0 + [0]
                [0 0 1]     [1],
   
                 [1 0 0]  
   [recip](x0) = [0 1 0]x0
                 [0 0 0]  ,
   
               [1 0 1]  
   [sqr](x0) = [0 1 0]x0
               [0 0 1]  ,
   
                 [1 1 1]     [1]
   [terms](x0) = [0 1 0]x0 + [0]
                 [0 0 0]     [1]
  orientation:
              [1 1 1]    [1]    [1 1 1]    [0]                      
   terms(N) = [0 1 0]N + [0] >= [0 1 0]N + [0] = cons(recip(sqr(N)))
              [0 0 0]    [1]    [0 0 0]    [1]                      
   
              [1]    [1]      
   sqr(0()) = [0] >= [0] = 0()
              [0]    [0]      
   
              [1]    [0]      
   sqr(s()) = [1] >= [1] = s()
              [1]    [1]      
   
              [1]    [1]      
   dbl(0()) = [0] >= [0] = 0()
              [1]    [0]      
   
              [0]    [0]      
   dbl(s()) = [1] >= [1] = s()
              [1]    [1]      
   
                [1 1 1]    [1]         
   add(0(),X) = [1 1 1]X + [1] >= X = X
                [1 1 1]    [1]         
   
                [1 1 1]    [1]    [0]      
   add(s(),Y) = [1 1 1]Y + [1] >= [1] = s()
                [1 1 1]    [1]    [1]      
   
                      [1]    [0]        
   first(0(),X) = X + [0] >= [0] = nil()
                      [0]    [0]        
   
                        [1 1 0]    [1]    [1 1 0]    [0]          
   first(s(),cons(Y)) = [0 1 0]Y + [1] >= [0 1 0]Y + [0] = cons(Y)
                        [0 0 1]    [1]    [0 0 1]    [1]          
  problem:
   sqr(0()) -> 0()
   dbl(0()) -> 0()
   dbl(s()) -> s()
  Matrix Interpretation Processor: dim=3
   
   interpretation:
                [1 0 0]     [1]
    [dbl](x0) = [0 0 0]x0 + [0]
                [0 0 0]     [0],
    
          [0]
    [s] = [0]
          [0],
    
          [0]
    [0] = [0]
          [0],
    
                [1 0 0]     [1]
    [sqr](x0) = [0 0 0]x0 + [0]
                [0 0 0]     [0]
   orientation:
               [1]    [0]      
    sqr(0()) = [0] >= [0] = 0()
               [0]    [0]      
    
               [1]    [0]      
    dbl(0()) = [0] >= [0] = 0()
               [0]    [0]      
    
               [1]    [0]      
    dbl(s()) = [0] >= [0] = s()
               [0]    [0]      
   problem:
    
   Qed