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=(−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.