Complex Sequence is Null iff Positive Integer Powers of Sequence are Null

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Theorem

Let $\sequence {z_n}_{n \mathop \in \N}$ be a complex sequence.

Let $k \in \N$.


Then:

$z_n \to 0$

if and only if:

$z_n^k \to 0$


Proof

Necessary Condition

Suppose that:

$z_n \to 0$

Let $\epsilon > 0$.

Then from the definition of convergent sequence we can find $N \in \N$ such that for $n \ge N$ we have:

$\cmod {z_n} < \epsilon^{1/k}$

then:

$\cmod {z_n}^k < \epsilon$

From Power of Complex Modulus equals Complex Modulus of Power, this gives:

$\cmod {z_n^k} < \epsilon$

for all $n \ge N$.

So:

$z_n^k \to 0$

$\Box$

Sufficient Condition

Suppose that:

$z_n^k \to 0$ for some $k$.

Let $\epsilon > 0$.

Then we can find $N \in \N$ such that:

$\cmod {z_n^k} < \epsilon^k$

From Power of Complex Modulus equals Complex Modulus of Power, we have:

$\cmod {z_n}^k < \epsilon^k$

so:

$\cmod {z_n} < \epsilon$

Since $\epsilon > 0$ was arbitrary, we have:

$z_n \to 0$

from the definition of a convergent complex sequence.

$\blacksquare$