Question

Exercise 3. Suppose that |2 < 2. Prove that the series converges absolutely.

Answer 1

Let $z_n=\frac{2z^2}{n^2&plus;|z|}$ , then to show $|z_n|=\frac{|2z^2|}{n^2&plus;|z|}$ is convergent.

Note that $|z_n|=\frac{|2z^2|}{n^2&plus;|z|}\le \frac{8}{n^2&plus;|z|}$ ---------------------**

Now note that $|z|\ge 0=> |z|&plus;n^2\ge n^2=> \frac{1}{n^2}\ge \frac{1}{|z|&plus;n^2}$

This gives us $|z_n|=\frac{|2z^2|}{n^2&plus;|z|}\le \frac{8}{n^2&plus;|z|}\le \frac{8}{n^2}$

Since $\{\sum \frac{8}{n^2}\}$ is convergent, by comparison test we get,  $\{\sum |z_n|\}$, thus $\{\sum z_n\}$ converges absolutely.

Feel free to comment if you have any doubts. Cheers!