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.