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.