0
$\begingroup$

Edited:

Let $\{z_n\}$ be a complex sequence with $\Re\{z_n\}\ge 0$. Suppose that $\sum_{k=0}^\infty z_n$ and $\sum_{k=0}^\infty z_n^2$ converge. Prove that $\sum_{k=0}^\infty z_n^2$ converges absolutely.

Proof:

Let $z=x+iy$. Then $\sum_{k=0}^\infty z_k= \sum_{k=0}^\infty x_k+i\sum_{k=0}^\infty y_k$, thus $\sum_{k=0}^\infty x_k$ and $\sum_{k=0}^\infty y_k$ converge. Similarly, $\sum_{k=0}^\infty z_k^2= \sum_{k=0}^\infty (x_k^2-y_k^2)+i\sum_{k=0}^\infty x_ky_k$

$=\sum_{k=0}^\infty (x_k-y_k)(x_k+y_k)+2i\sum_{k=0}^\infty |x_k|y_k=(\sum_{k=0}^\infty x_k-\sum_{k=0}^\infty y_k)(\sum_{k=0}^\infty x_k+\sum_{k=0}^\infty y_k)+2i\sum_{k=0}^\infty |x_k|y_k$

$=(\sum_{k=0}^\infty x_k)^2-(\sum_{k=0}^\infty y_k)^2+2i\sum_{k=0}^\infty |x_k|y_k = x^2 - y^2+2i\sum_{k=0}^\infty |x_k|y_k.$

We can draw from this that $\sum_{k=0}^\infty (x_k^2-y_k^2)=x^2 - y^2$, so $\sum_{k=0}^\infty x_k^2-\sum_{k=0}^\infty y_k^2=(\sum_{k=0}^\infty x_k)^2-(\sum_{k=0}^\infty y_k)^2$ (is this correct? I'm not sure).

Now, $\sum_{k=0}^\infty |z_k|^2=\sum_{k=0}^\infty (x_k^2+ y_k^2)$ converges, so $\sum_{k=0}^\infty z_n^2$ converges absolutely.

Do you think this proof might be correct?

$\endgroup$
  • $\begingroup$ In your edit there are now a bunch of mistakes in the algebra. You can't just pull the series into the multiplication, those steps are incorrect. My answer highlighted the flaw in your proof, and should be sufficient to conclude the proof. I will add a counter-example in a few minutes to show that if $\sum_{k=0}^{\infty} (x_k^2-y_k^2)$ does not converge absolutely then the series in question won't either. $\endgroup$ – adfriedman Feb 17 '17 at 7:35
1
$\begingroup$

Your "Similarly" step implicitly assumes you can break apart the real parts. This should actually be given as $$\sum_{k=0}^{\infty} z_k^2 = \sum_{k=0}^{\infty}(x_k^2 + 2ix_ky_k - y_k^2) = \sum_{k=0}^{\infty} (x_k^2-y_k^2) + 2i\sum_{k=0}^{\infty}x_ky_k $$

Because $\sum_{k=0}^{\infty} (x_k^2-y_k^2)$ might only be conditionally convergent, and for such cases $\sum_{k=0}^{\infty} x_k^2$ would not converge (hence $\sum_{k=0}^{\infty}|z_k^2|$ does not converge).

If you require $\mathfrak{R}\{z_n\}=x_n\geq 0$ then $\sum_{k=0}^{\infty} x_k$ is absolutely convergent and so $\sum_{k=0}^{\infty} x_k^2$ is convergent. Using properties guaranteed by this convergence, $\sum_{k=0}^{\infty}x_k^2 - \sum_{k=0}^{\infty} (x_k^2-y_k^2) = \sum_{k=0}^{\infty}y_k^2 < \infty$, hence $$ \sum_{k=0}^{\infty}|z_k^2| = \sum_{k=0}^{\infty} x_k^2 + \sum_{k=0}^{\infty} y_k^2 < \infty$$

Clearly requiring $\mathfrak{R}\{z_n\} \leq 0$ or $\mathfrak{I}\{z_n\}\geq 0$ or $\mathfrak{I}\{z_n\}\leq 0$ could also have worked in the proof.

$\endgroup$
  • $\begingroup$ Please check my edit. $\endgroup$ – sequence Feb 17 '17 at 7:26
  • $\begingroup$ Can you please clarify how $\sum_{k=0}^{\infty}x_k^2 - \sum_{k=0}^{\infty} (x_k^2-y_k^2) = \sum_{k=0}^{\infty}y_k^2$ exactly works? Did you separate out the $\sum_{k=0}^{\infty} (x_k^2-y_k^2)$ series? $\endgroup$ – sequence Feb 17 '17 at 7:33
  • 1
    $\begingroup$ If we know that $\sum_{k=0}^{\infty} z_k^2$ converges then both $\sum_{k=0}^{\infty} (x_k^2 - y_k^2) + 2i\sum_{k=0}^{\infty} x_ky_k$ are convergent. It is possible without further assumptions that $\sum_{k=0}^{\infty}x_k^2$ and $\sum_{k=0}^{\infty} y_k^2$ are both divergent. I showed that if $\mathfrak{R}\{z_k\} \geq 0$ then $\sum_{k=0}^{\infty} x_k^2$ is convergent. Using the fact that $\sum_{k=0}^{\infty}(x_k^2-y_k^2)$ also converges, we may safely combine sums term by term $\sum_{k=0}^{\infty}x_k^2 - \sum_{k=0}^{\infty} (x_k^2 - y_k^2) = \sum_{k=0}^{\infty}(x_k^2 -(x_k^2-y_k^2))$ $\endgroup$ – adfriedman Feb 17 '17 at 7:48
  • 1
    $\begingroup$ $=\sum_{k=0}^{\infty} y_k^2$. Therefore both $\sum_{k=0}^{\infty}x_k^2$ and $\sum_{k=0}^{\infty}y_k^2$ are both convergent, so $\sum_{k=0}^{\infty} |z_k^2|$ converges. Though the algebraic steps might seem a bit backwards, this is necessary in order to avoid using $\sum_{k=0}^{\infty} y_k^2$ before we show the expression even makes sense (i.e., is convergent). $\endgroup$ – adfriedman Feb 17 '17 at 7:50
  • $\begingroup$ Very cool, good to know. I was obviously lacking some theory in this area. Thanks for clearing that up. $\endgroup$ – sequence Feb 17 '17 at 8:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.