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:
- $z_n \to 0$
- $z_n^k \to 0$
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.
$\blacksquare$