I noticed something when scribbling the zeta function -
$$\zeta(3) = \sum_{n=1}^{\infty}\frac{1}{n^3} = 1 + \frac{1}{8} + \frac{1}{27} + \frac{1}{64} + \frac{1}{125} + \frac{1}{216} + \frac{1}{343} + \frac{1}{512} + \frac{1}{729} + \frac{1}{1000}\: +\: ...$$
and
$$\zeta(6)= \sum_{n=1}^{\infty}\frac{1}{n^6} = 1 + \frac{1}{64} + \frac{1}{729} + \frac{1}{4096} + \frac{1}{15625} + \frac{1}{46656} + \frac{1}{117649}\: +\: ...$$
Note that $64 = 2^6 = 4^3$, $729 = 3^6 = 9^3$, $4096=4^6=16^3$ ... because $$\zeta(6)= \sum_{n=1}^{\infty}\frac{1}{n^6}= \sum_{n=1}^{\infty}\frac{1}{(n^2)^3}$$
Substituting into $\zeta(3)$ produces
$$\zeta(3) = \zeta(6) + \Big(\frac{1}{8} + \frac{1}{27} + \frac{1}{125} + \frac{1}{216} + \frac{1}{343} + \frac{1}{512} + \frac{1}{1000}\: +\: ...\Bigr)$$
Every term that cancels is in the form of $4^3, 9^3, 16^3, 25^3, 36^3$. Thinking about the "perfect square" in reverse would generate the values of $2^6=64$, $3^6=729$, $4^6=4096$, $5^6=15625$ which correspond to the numbers above.
Is there any known summation formula for
$$\sum_{n=1}^{\infty}\frac{1}{n^3}$$
where n is not a perfect square such as 4, 9, 16, 25, 36, 49, 64 ....
I scanned the internet and found this and this.
Although, these don't seem to be of help. Is there a known summation formula for
$$\Big(\frac{1}{8} + \frac{1}{27} + \frac{1}{125} + \frac{1}{216} + \frac{1}{343} + \frac{1}{512} + \frac{1}{1000}\: +\: ...\Bigr)$$
which excludes all the perfect squares. Could I find this inside the OEIS?
If you doodle with $\zeta(3)$, you can also create
$$\zeta(3) = 1 + \frac{1}{8}\zeta(3) + \Bigl(\frac{1}{27} + \frac{1}{125} + \frac{1}{343} + \frac{1}{729} + \frac{1}{1331} + \frac{1}{2197}\: +\: ... \Bigr)$$
Although, this doesn't seem to help.
Edit: The Dirichlet eta function is defined as
$$\hspace{35mm}\eta(s) = \sum_{n=1}^{\infty}{(-1)^{n-1} \over n^s}\hspace{45mm}\forall\hspace{2mm}\Re(s) > 0$$
You can derive this relationship by (shown inside this post)
\begin{align*} \zeta(s)&=\sum_{n=1}^\infty \frac 1{n^s}\\ &=\sum_{n=1}^\infty \frac {(-1)^{n+1}}{n^s}+2\sum_{n=1}^\infty \frac 1{(2n)^s}\\ &=\sum_{n=1}^\infty \frac {(-1)^{n+1}}{n^s}+2^{1-z}\,\zeta(s)\\ \end{align*}
the end result is
$$\hspace{43mm}(1-2^{1-s})\zeta(s)=\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^s} \hspace{28mm}\forall\hspace{2mm}\Re(s) > 1$$
The answer below looks exactly like the multiplication theorem.
For
$$~\displaystyle\zeta(s)=\frac1{n^s}\sum_{k=1}^n\zeta\bigg(s,~\frac kn\bigg)~$$
This post is very similar, but has a slight variation. The answer for this post shows
We start with $\displaystyle \zeta(s,z) = \sum_{n=0}^{\infty} \frac{1}{(z+n)^{s}}$.
First, substitute in $mz$ for $z$:
$$\begin{align} \zeta(s,mz) &= \sum_{n=0}^{\infty} \frac{1}{(mz+n)^{s}} \\ &= \frac{1}{m^s}\sum_{n=0}^{\infty} \frac{1}{(z+\frac{n}{m})^{s}}. \end{align}$$
Now write $n = n'm + n''$, where now $n'$ will range from $0$ to $\infty $ and $n''$ will range from $0$ to $m-1$. Putting this above gives $$\begin{align} &= \frac{1}{m^s}\sum_{\substack{n'\geq 0 \\ n'' \bmod m}} \frac{1}{(z+\frac{n''}{m} + n')^{s}} \\ &= \frac{1}{m^s} \sum_{k = 0}^{m-1} \sum_{n' \geq 0} \frac{1}{(z + \frac km + n')^s} \\ &= \frac{1}{m^s}\sum_{k =0}^{m-1}\zeta\left(s,z+\frac{k}{m}\right), \end{align}$$
Set $z=0$ to get the desired form shown by the multiplication theorem.
