Whether the sequence is converges or diverges and obtain the limit if the sequence is converges.

The sequence is converges and its limit is −1.

Given:

The sequence is an=cos(nπn+1) .

Definition used:

If an is sequence and limn→∞an is exists then the sequence an is convergent otherwise it is divergent.

Results used:

(1) If f(x) is a function and c is any constant, then limx→a[c⋅f(x)]=c⋅limx→af(x) .

(2) If f(x) and g(x) are two functions, then limx→a[f(x)g(x)]=limx→af(x)limx→ag(x) and limx→ag(x)≠0 .

(3) If f(x) and g(x) are two functions, then limx→a[f(x)±g(x)]=limx→af(x)±limx→ag(x) .

(4) Limit chain rule:

If limu→bf(u)=L and limx→ag(x)=b with f(x) is continuous at x=b , then the value of limx→af(g(x)) is L.

Calculation:

Obtain the limit of the sequence to investigate whether the sequence is converges or diverges.

Compute limn→∞an=limn→∞cos(nπn+1) .

Let g(n)=nπn+1 and f(u)=cos(u) .

Obtain limn→∞g(n) .

limn→∞g(n)=limn→∞(nπn+1)=π⋅limn→∞(nn+1)            (By Result (1))

Divide the numerator and the denominator by the highest power.

limn→∞g(n)=π⋅limn→∞(nnn+1n)=π⋅limn→∞(11+1n)=π⋅limn→∞1limn→∞(1+1n)             (By Result (2))=π⋅1limn→∞(1)+limn→∞(1n)    (By Result (3))

Apply infinity property limn→∞(kna)=0 and simplify the numerator and the denominator.

limn→∞g(n)=π1+1∞  =π1+0=π1=π

Therefore, limn→∞g(n)=π .

Obtain limn→πf(u) .

limu→πf(u)=limu→π(cos(u))              =cos(π)              =−1

Therefore, limu→0f(u)=−1 .

Apply limit chain rule and obtain the limit.

Since f(u)=cos(u) and g(n)=nπn+1 , limn→∞cos(g(n))=−1 and limn→∞cos(nπn+1)=−1 .

Therefore, limn→∞f(g(n))=−1 .

Since limn→∞an is exist and by using the definition of the sequence, it can be concluded that the sequence is converges to the limit 1.

Since the sequence an exists as n tends to infinity, and by using the definition the sequence is convergent with limit is −1.

Therefore, the sequence is converges to the limit −1.