Does (1/n+1/ni) converge to zero?

[gottfried]

Homework Statement

Does the sequence $\frac{1}{n}$+$\frac{1}{n}$i converge to zero with respect to the metric d(z,w) = |z|+|w|.

The Attempt at a Solution

So I realize I want √($\frac{1}{n}$²+$\frac{1}{n}$²)=$\frac{√2}{n}$ to be less than ε. So for all n greater than $\frac{√2}{ε}$ the order $\frac{√2}{n}$<ε will hold.

So using the definition for all ε>0 there exists and n_ε=$\frac{√2}{ε}$ such that m>n implies $\frac{√2}{n}$<ε.

Is my technique correct? Is finding a function for n in terms of ε that always maintains the equality sufficient for a proof? If this is the case and ε is a real number how to we ensure that n is a natural number? Does n have to be natural number or can that condition just be put on m.

 

[Dick]

Homework Statement

Does the sequence $\frac{1}{n}$+$\frac{1}{n}$i converge to zero with respect to the metric d(z,w) = |z|+|w|.

The Attempt at a Solution

So I realize I want √($\frac{1}{n}$²+$\frac{1}{n}$²)=$\frac{√2}{n}$ to be less than ε. So for all n greater than $\frac{√2}{ε}$ the order $\frac{√2}{n}$<ε will hold.

So using the definition for all ε>0 there exists and n_ε=$\frac{√2}{ε}$ such that m>n implies $\frac{√2}{n}$<ε.

Is my technique correct? Is finding a function for n in terms of ε that always maintains the equality sufficient for a proof? If this is the case and ε is a real number how to we ensure that n is a natural number? Does n have to be natural number or can that condition just be put on m.

Looks ok. You generally choose n by just saying that it's an integer greater than $\sqrt{\frac{2}{ε}}$. No need to worry about whether $\sqrt{\frac{2}{ε}}$ itself is an integer.