Given:
The sequence is
an=ln(n+1)−ln(n)
.
Definition used:
If
an
is a sequence and
limn→∞an
exists, then the sequence
an
is said to be converges; otherwise it diverges.
Limit chain rule:
“If
limu→bf(u)=L
and
limx→ag(x)=b
with
f(x)
is said to be 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 converges or diverges.
Compute the value of
limn→∞an=limn→∞(ln(n+1)−ln(n))
.
Apply the Logarithm Quotient rule, then
logc(a)−logc(b)=logc(ab)
.
limn→∞(ln(n+1)−ln(n))=limn→∞(ln(n+1n))
Consider the functions
g(n)=n+1n
and
f(u)=ln(u)
.
Obtain
limn→∞g(n)
.
limn→∞g(n)=limn→∞(n+1n)
Divide numerator and the denominator by the highest power.
limn→∞(n+1n)=limn→∞(n+1nnn) =limn→∞(1+1n1) =limn→∞(1+1n)=limn→∞(1)+limn→∞(1n)
Apply infinity property
limn→∞(kna)=0
and simplify the terms of the expressions.
limn→∞(n+1n)=1+1∞=1+0 =1
Therefore,
limn→∞g(n)=1
.
Obtain
limu→1f(u)
.
limu→1f(u)=limu→1(ln(u)) =ln(1) =0
Apply the limit chain rule and obtain the limit.
Since
g(n)=n+1n
and
f(u)=ln(u)
,
limn→∞(ln(g(u)))=0
and
limn→∞(ln(n+1n))=0
.
Therefore,
limn→∞f(g(n))=0
.
Since
limn→∞an
exists and by using the definition of the sequence, it can be concluded that the sequence converges to the limit 0.
Therefore, the sequence converges to the limit 0.