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

Example of Divergent Real Sequence to Positive Infinity

Let $\alpha \in \Q_{>0}$ be a strictly positive rational number.

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

$a_n = n^\alpha$

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

Proof

We are to demonstrate that $n^\alpha \to +\infty$ as $n \to \infty$.

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

We need to find $N \in \N$ such that:

$\forall n > N: n^\alpha > H$

That is:

$\forall n > N: n > H^{1 / \alpha}$

We choose $N = H^{1 / \alpha}$.

The result follows.

$\blacksquare$

Sources

• 1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): $\S 4$: Convergent Sequences: $\S 4.28$: Example