Can I Define y and Explore Power Series Properties with This Sum?

[estro]

\mbox {Suppose I have: } \sum_{n=1}^\infty (\frac {x} {3})^{2n}

\mbox{Can I define } y= \frac {x} {3}

<br /> a_k(y) = \left\{<br /> \begin{array}{c l}<br /> (y)^k, &amp; \mbox{if } k= 2n\\<br /> \\<br /> (0)^k, &amp; \mbox{otherwise}<br /> \end{array}<br /> \right.<br />

\mbox {And then use all the cool properties of power series on } \sum_{k=1}^\infty a_k(y)

[I edited my question]

 

[Dickfore]

Of course, but the convergence properties you know are given in terms of y then. You need to translate them back in terms of x. Essentially, substitute y = (x/3)^{2} everywhere.

 

[HallsofIvy]

\mbox {Suppose I have: } \sum_{n=1}^\infty (\frac {x} {3})^{2n}

\mbox{Can I define } y= \frac {x} {3}

<br /> a_k(y) = \left\{<br /> \begin{array}{c l}<br /> (y)^k, &amp; \mbox{if } k= 2n\\<br /> \\<br /> (0)^k, &amp; \mbox{otherwise}<br /> \end{array}<br /> \right.<br />

\mbox {And then use all the cool properties of power series on } \sum_{k=1}^\infty a_k(y)

[I edited my question]

The simpler thing to do is write this as
\sum_{n=1}^\infty \left(\frac{x^2}{9}\right)^n
so it is a geometric series with "common ratio" of x^2/9.

 

[Gib Z]

\mbox {Suppose I have: } \sum_{n=1}^\infty (\frac {x} {3})^{2n}

\mbox{Can I define } y= \frac {x} {3}

<br /> a_k(y) = \left\{<br /> \begin{array}{c l}<br /> (y)^k, &amp; \mbox{if } k= 2n\\<br /> \\<br /> (0)^k, &amp; \mbox{otherwise}<br /> \end{array}<br /> \right.<br />

\mbox {And then use all the cool properties of power series on } \sum_{k=1}^\infty a_k(y)

[I edited my question]

Indeed you can, and that's how you can easily do problems without getting fooled like

\sum_{n=1}^{\infty} n^n z^{n^n}

 

[estro]

Thank you guys!