Can you transform the denominator of each term in an infinite series? Is there any way to transform the denominator in a power series?
For example,
$$\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6}$$ and $$\sum_{n=1}^{\infty}\frac{1}{(4n)^2}=\frac{\pi^2}{96}.$$ My question asks if you can convert each term in an infinite zeta series from $n$ to $an+b.$ As shown above, scaling is easily done however I cannot think of any way to add to each denominator. i.e.
$$ \sum_{n=1}^{\infty}\frac{1}{n^2} \to \sum_{n=1}^{\infty}\frac{1}{(an+b)^2}.$$
 A: The Wikipedia article Hurwitz zeta function states

In mathematics, the Hurwitz zeta function is one of the many
zeta functions. It is formally defined for complex variables $\,s\,$
with $\,\Re(s)>1\,$ and $\,a\ne 0,-1,-2,\dots\,$ by
$$\zeta(s,a) = \sum_{n=0}^\infty \frac{1}{(n+a)^{s}}.$$

You wrote

My question asks if you can convert each term in an infinite power series from $n$ to $an+b$.

In general, no. That is why the Hurwitz zeta function is needed to
generalize the Riemann zeta function. In some special cases, it is
possible to get results. The Wikipedia article states

If $\,a=1\,$ the Hurwitz zeta function reduces to the Riemann
zeta function itself; if $\,a=1/2\,$ it reduces to the Riemann
zeta function multiplied by a simple function of the complex
argument $\,s.$

As a first step, if $\,a\ne 0,\,$ it is easy to get
$$ \sum_{n=1}^\infty \frac1{(an+b)^s} = a^{-s}\zeta(s,b/a). $$
A simple result is that if $\,k>0\,$ is an integer, then
$$ \zeta(s,k) = \sum_{n=1}^\infty \frac1{(n+k)^s}
= \zeta(s) - \sum_{n=1}^k \frac1{n^s}. $$
A: If we have something like following
$$S=\sum_{n=a}^{b}p(n)$$
then $$\frac{S}{\alpha}=\frac{\sum_{n=a}^{b}p(n)}{\alpha}=\sum_{n=a}^{b}\frac{p(n)}{a}$$
For eg. In your case $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6}$, then $\displaystyle\frac{1}{16}\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{96}\implies \displaystyle\sum_{n=1}^{\infty}\frac{1}{16n^2}=\frac{\pi^2}{96} $
Therefore you can go from $\sum_{n=1}^{\infty}\frac{1}{n^2}$ to $\sum_{n=1}^{\infty}\frac{1}{(an)^2}$ but not in general to $\sum_{n=1}^{\infty}\frac{1}{(an+b)^2}$
