# $\lim_{ k \rightarrow \infty } { \frac{ \lambda^k }{k}} = \infty$ when $1 < |\lambda| \in \mathbb{C}$.

Can someone show why $\lim_{ k \rightarrow \infty } { \frac{ \lambda^k }{k}} = \infty$ when $1 < |\lambda| \in \mathbb{C}$.

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Just to make things a little easier to follow let $|\lambda|=1+x$ with $x >0$.

Then,

$$(1+x)^k \geq 1+ \binom{k}{1}x + \binom{k}{2}x^2> \frac{k(k-1)}{2}x^2 \,.$$

Thus

$$\left| \frac{\lambda^k}{k} \right| \ge \frac{k(k-1)x^2}{2}\frac{1}{k}=\frac{x^2}{2}(k-1)$$

Your conclusion follows immediately from here.

P.S. By exactly the same idea, or simply by Bernoulli, you can prove the following generalization:

If $a_n$ is a complex sequence, so that $\lim_n \left| \frac{a_{a+1}}{a_n} \right| =x >1$ then $\lim_n a_n =\infty$.

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One can use L'Hospitals rule, $$\lim \frac{\lambda^k}{k}=\lim \frac{\lambda^k \ln(\lambda)}{1}=\ln(\lambda)\cdot\lim \lambda^k=\infty$$ the last equality is because $|\lambda |>1$.

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You can use \ln and \cdot instead of ln and $.$ – The Chaz 2.0 Jul 9 '12 at 16:22
...and you can use \lim instead of \mbox{lim } – Michael Hardy Jul 9 '12 at 17:15

To show that a complex sequence $(z_n)$ is unbounded, we only need to show that the real sequence $(|z_n|)$ diverges to $+\infty$.

In the given sequence, $|\lambda^k/k|=|\lambda|^k/k$ and since the exponential growth dominates linear growth we will have that $\lim_{k \to +\infty}|\lambda^k/k|=+\infty$ as required.

Note: We can prove the claim rigorously without just vaguely claiming the exponential growth dominates linear growth. But I believe that we need to be sufficiently rigorous in mathematics not more, not less!

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