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

The sequence converges to the limit ln(2) .

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) .