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

The sequence converges to the limit 1.

Given:

Consider the given sequence as an=lnnln2n .

Definition used:

If an is a sequence and limn→∞an exists, then the sequence an is said to be converges; otherwise it diverges.

Laws of limits for sequences used:

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 .

If f(x) and g(x) are two functions, then, limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x) .

Calculation:

Obtain the limit of the sequence to investigate whether the sequence converges or diverges.

Compute limn→∞an=limn→∞ln(n)ln(2⋅n) .

Apply the Logarithm Product rule ln(a⋅b)=ln(a)+ln(b) .

limn→∞ln(n)ln(2⋅n)=limn→∞ln(n)ln(2)+ln(n)

Divide the numerator and the denominator by ln(n) .

limn→∞ln(n)ln(2⋅n)=limn→∞ln(n)ln(n)ln(2)+ln(n)ln(n)=limn→∞1ln(2)ln(n)+1

Apply the laws of limits for sequences and simplify the terms.

limn→∞ln(n)ln(2⋅n)=limn→∞(1)limn→∞(ln(2)ln(n)+1)=limn→∞(1)limn→∞(ln(2)ln(n))+limn→∞(1)=1limn→∞(ln(2)ln(n))+1

Then, apply the laws of limits for sequences in the denominator and simplify the terms.

limn→∞ln(n)ln(2⋅n)=1limn→∞(ln(2))limn→∞(ln(n))+1=1ln(2)∞+1=10+1=1

Since limn→∞an exists and by using the definition of the sequence, it can be concluded that the sequence converges to the limit 1.

Therefore, the sequence converges to the limit 1.