Limit of the Sequence kⁿ as n→∞

The limit of the sequence kⁿ as n→∞ depends on the value of k $$ \lim_{n \rightarrow ∞} k^n = \begin{cases} +∞ \:\:\: \text{if k>1} \\ 1 \:\:\: \text{if k=1} \\ 0 \:\:\: \text{if -1<k<1} \\ \text{does not exist} \:\:\: \text{if k \le -1} \end{cases} $$

Case k>1

$$ \lim_{n \rightarrow ∞} k^n $$

If k>1, we apply Bernoulli's inequality.

$$ k^n \ge 1+n(k-1) $$

Since k>1, the right-hand side diverges to +∞ as n tends to infinity.

$$ \lim_{n \rightarrow ∞} k^n \ge \lim_{n \rightarrow ∞} 1+n(k-1) $$

$$ \lim_{n \rightarrow ∞} k^n = +∞ $$

Hence, the sequence kⁿ diverges to +∞ as n→∞.

Note. For example, if k=1.1

This behavior is consistent with the corresponding exponential function f(x)=k^x.

 

 

Case k=1

$$ \lim_{n \rightarrow ∞} k^n $$

$$ \lim_{n \rightarrow ∞} 1^n $$

For every n, we have 1ⁿ=1.

Therefore, the sequence is constant and converges to 1.

$$ \lim_{n \rightarrow ∞} 1^n = 1 $$

Example. The sequence converges to 1 as n tends to infinity.

Case -1<k<1 with k≠0

$$ \lim_{n \rightarrow ∞} k^n $$

We rewrite the expression in an equivalent form

$$ \lim_{n \rightarrow ∞} \left( \frac{1}{ \frac{1}{k} } \right)^n $$

$$ \lim_{n \rightarrow ∞} \frac{1}{\left( \frac{1}{k} \right)^n} $$

The inequality

$$ -1 < k < 1 $$

is equivalent to

$$ |k| < 1 $$

Dividing both sides by |k|, with k≠0, yields

$$ \frac{|k|}{|k|} < \frac{1}{|k|} $$

$$ 1 < \frac{1}{|k|} $$

that is,

$$ \frac{1}{|k|} > 1 $$

Hence, the base in the denominator exceeds 1, and therefore

$$ \left( \frac{1}{|k|} \right)^n \rightarrow +∞ \quad \text{as } n \rightarrow ∞ $$

It follows that

$$ \lim_{n \rightarrow ∞} \frac{1}{\left( \frac{1}{k} \right)^n} = 0 $$

Thus, the sequence converges to 0 as n→∞.

Example. If k=0.9, the sequence converges to 0.

Case k=0

$$ \lim_{n \rightarrow ∞} k^n $$

$$ \lim_{n \rightarrow ∞} 0^n $$

For every n≥1, we have 0ⁿ=0.

Therefore, the sequence is constant and converges to 0.

$$ \lim_{n \rightarrow ∞} 0^n = 0 $$

In this case, the sequence is infinitesimal.

Example. If k=0, the sequence converges to 0.

Case k<-1

$$ \lim_{n \rightarrow ∞} k^n $$

If k<-1, the sequence diverges.

Its terms alternate in sign, while their absolute values diverge to +∞.

More precisely,

$$ \lim_{n \rightarrow ∞} k^n = \begin{cases} +∞ \:\: \text{if n is even} \\ -∞ \:\: \text{if n is odd} \end{cases} $$

Therefore, the limit does not exist as n tends to infinity.

$$ \lim_{n \rightarrow ∞} k^n = \text{does not exist} $$

Example. If k=-1, the sequence oscillates between 1 and -1, so the limit does not exist as n→∞.

If k=-1.1, the sequence diverges to +∞ when n is even and to -∞ when n is odd.

And so on.