NO

Problem 1: 

(VAR v_NonEmpty:S N:S X:S Y:S Z:S)
(RULES
add(0,X:S) -> X:S
add(s(X:S),Y:S) -> s(add(X:S,Y:S))
dbl(0) -> 0
dbl(s(X:S)) -> s(s(dbl(X:S)))
first(0,X:S) -> nil
first(s(X:S),cons(Y:S,Z:S)) -> cons(Y:S,first(X:S,Z:S))
half(dbl(X:S)) -> X:S
half(0) -> 0
half(s(0)) -> 0
half(s(s(X:S))) -> s(half(X:S))
sqr(0) -> 0
sqr(s(X:S)) -> s(add(sqr(X:S),dbl(X:S)))
terms(N:S) -> cons(recip(sqr(N:S)),terms(s(N:S)))
)

Problem 1: 

Dependency Pairs Processor:
-> Pairs:
 ADD(s(X:S),Y:S) -> ADD(X:S,Y:S)
 DBL(s(X:S)) -> DBL(X:S)
 FIRST(s(X:S),cons(Y:S,Z:S)) -> FIRST(X:S,Z:S)
 HALF(s(s(X:S))) -> HALF(X:S)
 SQR(s(X:S)) -> ADD(sqr(X:S),dbl(X:S))
 SQR(s(X:S)) -> DBL(X:S)
 SQR(s(X:S)) -> SQR(X:S)
 TERMS(N:S) -> SQR(N:S)
 TERMS(N:S) -> TERMS(s(N:S))
-> Rules:
 add(0,X:S) -> X:S
 add(s(X:S),Y:S) -> s(add(X:S,Y:S))
 dbl(0) -> 0
 dbl(s(X:S)) -> s(s(dbl(X:S)))
 first(0,X:S) -> nil
 first(s(X:S),cons(Y:S,Z:S)) -> cons(Y:S,first(X:S,Z:S))
 half(dbl(X:S)) -> X:S
 half(0) -> 0
 half(s(0)) -> 0
 half(s(s(X:S))) -> s(half(X:S))
 sqr(0) -> 0
 sqr(s(X:S)) -> s(add(sqr(X:S),dbl(X:S)))
 terms(N:S) -> cons(recip(sqr(N:S)),terms(s(N:S)))

Problem 1: 

Infinite Processor:
-> Pairs:
 ADD(s(X:S),Y:S) -> ADD(X:S,Y:S)
 DBL(s(X:S)) -> DBL(X:S)
 FIRST(s(X:S),cons(Y:S,Z:S)) -> FIRST(X:S,Z:S)
 HALF(s(s(X:S))) -> HALF(X:S)
 SQR(s(X:S)) -> ADD(sqr(X:S),dbl(X:S))
 SQR(s(X:S)) -> DBL(X:S)
 SQR(s(X:S)) -> SQR(X:S)
 TERMS(N:S) -> SQR(N:S)
 TERMS(N:S) -> TERMS(s(N:S))
-> Rules:
 add(0,X:S) -> X:S
 add(s(X:S),Y:S) -> s(add(X:S,Y:S))
 dbl(0) -> 0
 dbl(s(X:S)) -> s(s(dbl(X:S)))
 first(0,X:S) -> nil
 first(s(X:S),cons(Y:S,Z:S)) -> cons(Y:S,first(X:S,Z:S))
 half(dbl(X:S)) -> X:S
 half(0) -> 0
 half(s(0)) -> 0
 half(s(s(X:S))) -> s(half(X:S))
 sqr(0) -> 0
 sqr(s(X:S)) -> s(add(sqr(X:S),dbl(X:S)))
 terms(N:S) -> cons(recip(sqr(N:S)),terms(s(N:S)))
-> Pairs in cycle:
 TERMS(N:S) -> TERMS(s(N:S))

The problem is infinite.