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

The sequence converges to the limit 0.

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.