Prove if the sequence is convergent or divergent nn 1 cosn

Prove if the sequence is convergent or divergent {n/n + 1 cos(n pi)}_n = 1^infinity {ln n/n + 1}_n = 2^infinity {sin^2(n0/2^n}_n = 1^infinity

Solution

Let U(n) = {n/(n+1)} cosn and let v(n) = n/(n+1)

Since -1 cosn 1, U(n) V(x) and as such, if V(x) is convergent, U ( x) is also convergent.

Now Lim _{ntends to} n/n+1 = Lim _{n tends to} 1 /( 1 + 1/n) ( dividing Numerator & Denominator by n)

= 1/(1+0) = 1 (finite)

So V(n) is convergent and as U(n) is also convergent.

(b) Here let U(n) = logn / (n+1) and V(n) = 1/(n+1)

Lim _{n tends to} 1/n+1 = Lim _{n tends to} 1/n /( 1 + 1/n) = 0/0+1 = 0

So V(n) is convergent and as such U(n) is convergent.

(c) Here also U(n) = sin²n/2ⁿ and V(n) = 1/2ⁿ

^Since -1Sin²n1, U(n) V(x) and as such, if V(x) is convergent, U ( x) is also convergent.

^Here Lim _{n tends to} 1/2ⁿ = 0

So, V(x) is convergent and as such U(n) is convergent.