Reindexing a power series with substitution y = x/3

[estro]

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

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

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

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

[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]

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

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

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

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

[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]

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

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

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

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

[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!