Given:
The sequence is
an=(−1)n2n
.
Definition used:
If
an
is a sequence and
limn→∞an
exists, then the sequence
an
is said to be converges otherwise it diverges.
Theorem used:
If
limn→∞|an|=0
, then
limn→∞an=0
. (1)
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)
is a function and c is any constant, then,
limx→a[c⋅f(x)]=c⋅limx→af(x)
.
(3) If
f(x)
is a functions then,
limx→a[f(x)]b=[limx→af(x)]b
.
Calculation:
Obtain the limit of the sequence to investigate whether the sequence converges or diverges.
Compute
limn→∞an=limn→∞(−1)n2n
. (2)
The value of
(−1)n2n
is,
|(−1)n2n|=|(−1)n||2n|={|1||2n|if n is even|−1||2n|if n is odd=12n
The absolute value of
(−1)nn
when
n→∞
is obtained is as follows:
limn→∞|(−1)n2n|=limn→∞12n=limn→∞(1)limn→∞(2n) (By Result (1))=12⋅limn→∞(n)12 (By Result (2))=12⋅(limn→∞n)12 (By Result (3))
Apply the infinity property
limn→∞n=∞
and simplify the expressions.
limn→∞|(−1)n2n|=12⋅(∞)12=1∞=0
Therefore,
limn→∞|(−1)n2n|=0
Since,
limn→∞|(−1)n2n|=0
and by using Theorem (2), the value of
limn→∞(−1)n2n=0
.
Therefore,
limn→∞(−1)n2n=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 convergent to the limit 0.