Question

Show that if
∑ Zn is convergent , then ∑ Zn ^ 2 is also convergent

please more than method

Answer 1

Answer on Question #50127 - Math - Complex Analysis

Show that if $\sum z_{n}$ is convergent, then $\sum z_{n}^{2}$ is also convergent

Solution.

It cannot be shown, because it is false.

For example, let's consider series $\sum z_{n} = \sum \frac{(-1)^{n}}{n^{\frac{1}{2}}}$. Since $\frac{1}{n^{\frac{1}{2}}}$ positive and decreasing to the 0, so by the Leibniz's theorem, the original series is convergent.

Now consider series $\sum z_{n}^{2} = \sum \left(\frac{(-1)^{n}}{n^{\frac{1}{2}}}\right)^{2} = \sum \frac{1}{n^{\frac{1}{2}}}$. By the Integral Test, the improper integral

$$\int_{1}^{\infty} \frac{dx}{x^{1/2}} = \lim_{A \to \infty} \int_{1}^{A} \frac{1}{x^{\frac{1}{2}}} dx = \lim_{A \to \infty} \left(2x^{\frac{1}{2}}\right)_{1}^{A} = \infty \quad \text{does not exist, therefore the series } \sum \frac{1}{n^{\frac{1}{2}}} \text{ is divergent.}$$

This means that the original statement is not true.

Nevertheless, there exist some cases when the proposition is true.

For example, if we change our proposition like that: $z_{n}$ are real, positive numbers and $\sum z_{n}$ is convergent, then $\sum z_{n}^{2}$ is also convergent.

Let's prove this variant of statement.

Since $\sum z_{n}$ is convergent, then by necessary condition, $\lim_{n\to \infty}\left|z_n\right| = 0$. That is why there exists a natural number $N$ such that $\forall n > N\Rightarrow |z_n| < 1$. Then $\forall n > N\Rightarrow 0\leq |z_n^2| = |z_n|^2\leq |z_n|$, so from comparison test we obtain that $\sum z_{n}^{2}$ is convergent.

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