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

The sequence converges to the limit −2.

Given:

The sequence is an=n−n+1n+3 .

Definition used:

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

Results used:

(1) 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 .

(2) 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.

Simplify an=n−n+1n+3 as follows:

an=n−n+1n+3=n−n⋅n+3⋅n+1⋅n+1⋅3              (Radical Rule: ab=ab)=n−n2+3n+n+3=n−n2+4n+3

Multiply by the conjugate of n−n2+4n+3 .

n−n+1n+3=n−n2+4n+3⋅n+n2+4n+3n+n2+4n+3=(n−n2+4n+3)⋅(n+n2+4n+3)n+n2+4n+3=n2−(n2+4n+3)2n+n2+4n+3 [Since (a−b)(a+b)=a2−b2]=n2−(n2+4n+3)n+n2+4n+3

Simplify the terms as shown below:

n−n+1n+3=n2−n2−4n−3n+n2+4n+3=−4n−3n+n2+4n+3

Therefore, an=−4n−3n+n2+4n+3 .

Compute limn→∞an=limn→∞−4n−3n+n2+4n+3

Divide the numerator and the denominator by n.

limn→∞−4n−3n+n2+4n+3=limn→∞−4n−3nn+n2+4n+3n   =limn→∞−4nn−3nnn+n2+4n+3n   =limn→∞−4−3n1+n2+4n+3n2   =limn→∞−4−3n1+1+4n+3n2

Apply the Results (1) and (2) and obtain the value of limit.

limn→∞−4n−3n+n2+4n+3=limn→∞(−4−3n)limn→∞(1+1+4n+3n2)   =limn→∞(−4)−limn→∞(3n)limn→∞(1)+(limn→∞(1+4n+3n2))12   =−4−limn→∞(3n)1+(limn→∞(1)+limn→∞(4n)+limn→∞(3n2))12

Apply infinity property limn→∞(kna)=0 and simplify the terms.

limn→∞−4n−3n+n2+4n+3=−4−01+(1+0+0)12=−4−01+(1)12=−42=−2

Since limn→∞an exists and by the definition of the sequence, it can be concluded that the sequence converges to the limit −2.

Therefore, the sequence converges to the limit −2.