Given:
The sequence is
an=ln(2n2+1)−ln(n2+1)
.
Definition used:
If
an
is sequence and
limn→∞an
exists, then the sequence
an
is said to be converges otherwise it is 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
limn→∞(ln(2n2+1)−ln(n2+1))
.
Apply the Logarithm Quotient rule,
logc(a)−logc(b)=logc(ab)
.
limn→∞(ln(2n2+1)−ln(n2+1))=limn→∞(ln(2n2+1n2+1))
Consider the functions
g(n)=2n2+1n2+1
and
f(u)=ln(u)
and obtain
limn→∞g(n)
.
limn→∞g(n)=limn→∞(2n2+1n2+1)
Divide numerator and the denominator by the highest power.
limn→∞(2n2+1n2+1)=limn→∞(2n2+1n2n2+1n2) =limn→∞(2+1n21+1n2) =limn→∞(2+1n2)limn→∞(1+1n2) =limn→∞(2)+limn→∞(1n2)limn→∞(1)+limn→∞(1n2)
Apply infinity property
limn→∞(kna)=0
and simplify the terms of the expressions.
limn→∞(2n2+1n2+1)=2+1∞1+1∞ =2+01+0 =21 =2
Obtain
limu→2f(u)
.
limu→2f(u)=limu→2(ln(u)) =ln(2)
Apply the limit chain rule and obtain the limit.
Since
g(n)=2n2+1n2+1
and
f(u)=ln(u)
, the value of
limn→∞f(2n2+1n2+1)=ln(2)
and
limn→∞(ln(2n2+1n2+1))=ln(2)
.
Therefore,
limn→∞(f(g(n)))=ln(2)
.
Since
limn→∞an
exists and by using the definition, it can be concluded that the sequence converges to the limit
ln(2)
.
Therefore, the sequence converges to the limit
ln(2)
.