Question

prove that if the series \[\sum_{n=1}^{\infty}z _{n}\] converges absolutley, so does the series
\[\sum_{n=1}^{\infty}z _{n}^2\]

Answer 1

z is complex number

Answer 2

Maybe like this?
\[(x+iy)^{2} \le |z|^{2} \]
and since Sigma|z| converges so would the Sigma|z|2

Answer 3

but the left and right hand term are not equal .. the left hand term is a complex no while the right hand term a real number

Answer 4

sry forgot the module sign. I ment their absolut value

Answer 5

\[|(x+iy)^{2}|\le |z|^{2}\]

Answer 6

would be equal if x=y

Answer 7

for \(n\) large enough (\(n>N\)) we have \(|z_n|<1\)

then

\(|z_n^2|<|z_n|\) for all \(n>N\)

Answer 8

good point, thx

Answer 9

actualy really nice and easy

Answer 10

yep