Divergent Real Sequence to Positive Infinity/Examples/2^n

Example of Divergent Real Sequence to Positive Infinity

Let $\sequence {a_n}_{n \mathop \ge 1}$ be the real sequence defined as:

$a_n = 2^n$

Then $\sequence {a_n}$ is divergent to $+\infty$.

Proof 1

Let $H \in \R_{>0}$ be given.

By Boundedness of Nth Powers, $\sequence {a_n}$ is unbounded above.

Hence:

$\exists N \in \N: 2^N > H$

Then if $n > N$, we have:

$2^n > 2^N > H$

and the result follows.

$\blacksquare$

Proof 2

Let $H \in \R_{>0}$ be given.

We have that:

$2^n > n$

so setting $N = H$:

$\exists N \in \N: 2^n > H$

for all $n > N$.

$\blacksquare$

Sources

• 1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): $\S 4$: Convergent Sequences: Exercise $\S 4.29 \ (3) \ \text{(i)}$